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Stoichiometry
By: Ms. Michelle BurokerScott High School
Taylor Mill, KentuckyGo Eagles!
Let’s Get Started!
A Brief Introduction….A Brief Introduction….
Stoichiometry is the Stoichiometry is the mathematics of chemical mathematics of chemical reactions. This is a tutorial reactions. This is a tutorial meant to introduce you to meant to introduce you to the basics so that you will be the basics so that you will be able to calculate with able to calculate with confidence! confidence!
There is a brief quiz at the There is a brief quiz at the end of the lesson, you need end of the lesson, you need to take it, print out the to take it, print out the certificate at the end, and certificate at the end, and bring it to me for credit. bring it to me for credit.
Here We Go!
Have you ever wondered when you were watching fireworks exactly how they make that happen?
It’s Stoichiometry!
Stoichiometry is the study of quantitative relationships between amounts of reactants used and products formed by a chemical reaction. The purpose of this learning module is for you to interactively learn Stoichiometry: mathematical operations which are very important to scientists.
When you follow the lessons laid out in this assignment, hopefully you will gain an understanding of the process by which you can solve questions such as: if I start with so many grams of reactants, how many grams of product will I get?
Main Menu
Main MenuIf at any time, you need to return to the Main Menu, click on the “home”
icon at the bottom left of each screen.
The Mole (a.k.a. not the animal)
Mole to Mole Relationships
Mass to Mass Relationships
Limiting Reagents
Percent Yield
Are you ReadyTo Begin?!?
Return to the Beginning of
The SIM
The Mole
The SI base unit used to measure the amount of a substance!
One mole of anything is 6.02 x 1023 particles …– Remember that a particle can be: atoms, molecules, or formula units
Ex: 1 mole of C = 6.02 x 1023 atoms 1mole of NaCl = 6.02 x 1023 formula units
1 mole of H2O = 6.02 x 1023 molecules
The mass of any pure substance is equal to it’s molar mass
- Remember that the molar mass of any element is equal to its atomic weight and the molar mass of any molecule or formula unit is equal to the sum of the atomic weight’s of the individual atoms
- Ex: 1 mole of Ca weighs 40.08g
1 mole of H2O weighs 18.0g
Conversion Factors
Click HereTo Watch a Video about
The Mole
Conversion Factors
• The following are the conversion factors you need to “do the math:”
1.) To go from moles to grams:
number of moles given X atomic weight or molar mass
1 1 mole
2.) To go from grams to moles:
number of grams given X 1 mole
1 atomic weight or molar mass
Let’s Try Some Practice Problems
Conversion Factors
• The following are the conversion factors you need to “do the math:”
1.) To go from moles to grams:
number of moles given X atomic weight or molar mass
1 1 mole
2.) To go from grams to moles:
number of grams given X 1 mole
1 atomic weight or molar mass
Let’s Try Some Practice Problems
Let’s Practice!!!!
#1: How much does 5 moles of AgNO3
weight?
* Click on your choice!A. 169.91g
B. 849.40g
C. 339.76g
Try Again …
• Take a look at the conversation factor page again … use those to help you.
ConversationFactor Page
Back to theProblem
Try Again …
• Take a look at the conversation factor page again … use those to help you.
ConversationFactor Page
Back to theProblem
Let’s Practice!!!
#2. How many moles are 250g of NaCl?
*Click on your choice!
A. 58.45 moles
B. 1.46 x 104moles
C. 4.28moles
Mole to Mole RelationshipsNow that you have
mastered basic conversions between moles and grams … you’re ready to move on to calculations within chemical reactions!
The mole is how we relate reactants to other reactants … products to other products … or reactants to products. It’s called the Mole Ratio!
Say What???
The Mole RatioMole ratios are used as conversion factors to convert
a known number of moles of one substance to moles of another substance in the same chemical reaction.
Example:
2K(s) + 2H2O(l) 2KOH(aq) + H2(g)
For every 2 moles of potassium, you have 2 moles of H2O, so the mole ratio would look like this:
2 moles K 2 moles H2O
Let’s Try APractice Problem!
Let’s Practice!Let’s Practice!One disadvantage to One disadvantage to
burning propane (Cburning propane (C33HH88) is ) is that carbon dioxide (COthat carbon dioxide (CO22) ) is one of the products. is one of the products. The release of carbon The release of carbon dioxide increases the dioxide increases the growing concentration of growing concentration of COCO22 in the atmosphere. in the atmosphere. How many moles of How many moles of carbon dioxide are carbon dioxide are produced when 10.0 produced when 10.0 moles of propane are moles of propane are burned in excess oxygen burned in excess oxygen in a gas grill?in a gas grill? Check
Your Answer
Having Trouble Getting Started and
Need a Hint …
Here’s a Hint ….Here’s a Hint ….
You need to write a balanced chemical You need to write a balanced chemical equation before you can solve this equation before you can solve this
problem! problem!
Hopefully, you recognize this as a Hopefully, you recognize this as a combustion reaction. combustion reaction.
CC33HH88 + 5O + 5O22 3CO 3CO22 + 4H + 4H22OOOn To The
Answer!
Answer:
Your first step is to write the balanced chemical equation for the problem:
C3H8 + 5O2 3CO2 + 4H2O
If you start with 10.0moles of C3H8, then…(10.0moles C3H8) X (3molesCO2/1mole C3H8)= 30.0moles CO2
You needed the mole ratio between C3H8 and CO2 to solve the problem!
Moving Right Along!
Mass to Mass Relationships
• The coefficients of a balanced equation give the relative amounts (in moles) of reactants and products. Calculations to find the masses of materials involved in reactions are called Mass-Mass Problems.
Okay … buthow do I actuallysolve a problem?
The Pathway …
• There is a step- by- step method to solving mass- mass problems; the following steps must be followed!
Grams Reactant Moles Reactant Moles Product Grams Product
Convert Grams to Moles Mole Ratio Convert Moles to grams
*Note: If you begin with grams of Product, the process is simply done in reverse!
Let’s Look at an Example
Mass-Mass … An Example!How many grams of silver chloride can be produced from the reaction of 17.0g
of silver nitrate with excess sodium chloride solution?
Step 1: Write the chemical equation for the reaction.AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)
Step 2: Convert grams of reactant given to moles.(17.0g AgNO3) X (1mole/ 169.88gAgNO3)= 0.100moles AgNO3
Step 3: Using the mole relationship, convert moles of reactant into moles of product.
(0.100moleAgNO3) X (1mole AgNO3/ 1mole AgCl)= 0.100moles AgCl
Step 4: Convert moles to grams of product.(0.100moles AgCl) X (143.32g/1moleAgCl)= 14.33g AgCl
Ready to try oneOn your own??
Hmmm … Let’s See How You Do On Your Own.
Ammonium nitrate (NH4NO3), an important fertilizer, produces N2O
gas and H2O when it decomposes.
Determine the mass of water produced from the decomposition of
25.0g of solid ammonium nitrate.
Try it on your own before checking the answer!
Answer
The Origins of Fertilizer???
Fertilizer originated from a process known as the Haber Process, developed during WW I in Germany.
(If you’re interested in reading a bit more about the Haber Process … click the link for some brief info) …
Before you see the answer, follow these steps to solve the problem…
Step 1: Write the balanced chemical equation for the reaction.Step 2: Convert grams of reactant given to moles.Step 3: Using the mole relationship, convert moles of reactant into moles of product.Step 4: Convert moles to grams of product.
Now Take aLook at the
Answer
Okay … This Time it Really is The Answer!
NH4NO3 N2O + 2H2O
Convert grams of NH4NO3 to moles of NH4NO3.(25.0gNH4NO3/1) X (1mol NH4NO3/ 80.06gNH4NO3 )= 0.312mol NH4NO3
Convert moles of NH4NO3 to moles of H2O.
(0.312mol NH4NO3 /1) X (2mole H2O/1mole NH4NO3)= 0.624 mole H2O
Covert moles of H2O to grams of H2O .
(0.624molH2O/1) X (18.0g H2O/ 1mole H2O)= 11.24g H2O
Why do reactions stop?
The answer is … limiting reagents!
If you put 10 boys and 6 girls in a room and asked them to pair up: one boy to one girl, who will be left without a
partner?
The boys … there will be four boys left with no partner. The girls are the limiting factor, in other words, they
control how many pairs will be formed.
Keep Moving!
Limiting Reagents and Reactions
Just like the previous example, in the lab,
reactions are limited by the reactant present in the LEAST amount. So, to solve limiting
reagent problems, you have to first find out which reactant is the limiting reagent!!!
Hint … convert grams of reactant to moles!
Let’s Take a LookAt an Example!
Air Bags ….
The reaction between solid sodium and iron (III) oxide is one in a series of reactions that inflates an automobile
airbag.
6Na(s) + Fe2O3(s) 3Na2O(s) + 2Fe(s)
If 100.0g Na and 100.0g Fe2O3 are used in this reaction, determine:a. the limiting reagentb. the excess reagentc. the mass of solid iron producedd. the amount of excess reaction remaining after the reaction is
complete.
Take a look at How to work it out!
Watch AirBag
Deployment
The Solution …6Na(s) + Fe2O3(s) 3Na2O(s) + 2Fe(s)
1.) Determine the Limiting Reagent by Converting grams of product into moles of product.
(100.0gNa/1) X (1mol Na/ 23.0gNa)= 4.35mol Na
(100.0gFe2O3/1) X (1molFe2O3/159.7gFe2O3)= 0.6262mol Fe2O3
Fe2O3 is present in the least amount, therefore, it is the limiting reagent. The excess reactant would be Na.
2.) Determine the amount of solid Fe produced by converting moles of the limiting reagent into grams of product.
(0.6262mol Fe2O3/1) X (2mol Fe/1mol Fe2O3) X (55.85g Fe/1mol Fe)= 69.95g Fe
3.) To determine the amount of excess reactant left, convert moles of limiting reagent to grams of excess reactant and subtract that number from what you started with.
(0.6262mol Fe2O3/1) X (6mol Na/1mol Fe2O3) X (23.0g Na/ 1mol Na)= 86.41g Na
100.0g Na Started with – 86.41g Na used= 13.59g Na left in excessTry a Problem onYour Own Now …
Time to Try a Problem on Your Own! Good Luck
If 4.44g of calcium oxide are mixed with 7.77g of water, how many
grams of calcium hydroxide will form?
Try working it out on your own before
looking at the answer!!!
Answer
Do You Need
A Hint?
Need a Hint?
You need to write a balanced chemical
equation to solve this problem!
Hopefully you recognized this as a synthesis reaction
CaO + H2O Ca(OH)2
On To The Answer!
Answer
Begin by writing a balanced chemical equation for the reaction:
CaO + H2O Ca(OH)2
Determine the limiting reagent:(4.44g CaO/1) X (1mol CaO/ 56.08g CaO)= 0.0792mol CaO
(7.77g H2O/1) X (1mol H2O/ 18.0g H2O)= 0.432mol H2O
Convert moles of limiting reagent into grams of product:
(0.0792mol CaO/ 1) X (1mol Ca(OH)2/ 1mol CaO) X (74.10gCa(OH)2/1mol Ca(OH)2)= 5.88g Ca(OH)2
Almost Done!
How accurate Was my Experiment?
To answer this question … you need to know about the percent yield.
Percent yield is essentially how much product you got from a chemical reaction
vs. how much you should have gotten back according to your Stoichiometry
calculation.The Formula
Percent Yield
Percent Yield = Actual Yield X100
Theoretical Yield
Actual Yield= What you get from the experiment or the “experimental value”
Theoretical Yield= How much product the stoichiometric calculation says you should get, based on how much reactant you started with.
Let’s see how to useThis equation …
Here’s An Example ….
When potassium chromate (K2CrO4) is added to a solution containing 0.500g silver nitrate (AgNO3), solid silver chromate (Ag2CrO4) is formed.
A.) Determine the theoretical yield of silver chromate precipitate.
B.) Determine the percent yield if 0.455g silver chromate was actually recovered.
*If you follow the steps … find the theoretical yield first, then the calculation becomes quite simple.
Click Here if youWant to see the
Problem Worked Out
Answer to The Example ….
1.) As always, begin by writing a balanced chemical equation for the reaction:
K2CrO4 + 2AgNO3 Ag2CrO4 + 2KNO3
2.) Figure out the theoretical yield by converting 0.500g AgNO3 to grams of Ag2CrO4.
(0.500gAgNO3/1) X (1mol AgNO3/169.91g AgNO3)= 0.00294mol AgNO3
(0.00294mol AgNO3/1) X (1mol Ag2CrO4/ 2mol AgNO3)= 0.00147mol Ag2CrO4
(0.00147mol Ag2CrO4/1) X (331.8gAg2CrO4/ 1mol Ag2CrO4)= 0.488g Ag2CrO4
3.) Now that you have the theoretical yield, 0.488g Ag2CrO4, you can use the actual yield of 0.455g Ag2CrO4 and find the percent yield:
(0.455gAg2CrO4/ 0.488g Ag2CrO4) X 100= 93.2%
Yeah!!!
Great Job!!!!!!!Great Job!!!!!!!
Finally … you have completed this learning Finally … you have completed this learning module over Stoichiometry. My hope for you is module over Stoichiometry. My hope for you is
that you learned something! that you learned something!
The last leg of your journey is to complete a The last leg of your journey is to complete a questionnaire over this Learning Module. You questionnaire over this Learning Module. You
need to Print out the page, answer the need to Print out the page, answer the questions, and then turn the sheet into me for questions, and then turn the sheet into me for
credit.credit. Click Here to
Proceed to the Evaluation
The Evaluation
You must take this evaluation in order to receive the extra credit
for completing this Learning Module. Follow the evaluation link, print the page, answer the questions, and turn it into me!
The EvaluationThe End …
Thanks for StoppingBy
Correct!
Great Job!!!
Moving On!
Way To Go!!!!!!
Ready for What’s Next???
ReferencesReferences
Thank You and I hope you enjoyed the Experience!Thank You and I hope you enjoyed the Experience!Unless designated otherwise, all pictures are from clip Unless designated otherwise, all pictures are from clip
art.art.
Slide 1Slide 1 Scientist Picture:Scientist Picture:((http://istockphoto.com/imageindex/310/7/310758/The_Mad_Scientist.htmlhttp://istockphoto.com/imageindex/310/7/310758/The_Mad_Scientist.html ))Eagle Picture:Eagle Picture:((http://www.scott.kenton.kyschools.us/default1.htmhttp://www.scott.kenton.kyschools.us/default1.htm))
Slide 5Slide 5Mole Picture:Mole Picture:(http://www.ringwood.hants.sch.uk/school2/_Subjects/sciChemistry_ks3/(http://www.ringwood.hants.sch.uk/school2/_Subjects/sciChemistry_ks3/SpencerNewSite/page_7E_acids.html)SpencerNewSite/page_7E_acids.html)
Resource Book Used:Resource Book Used:Dingrando, Laurel, Kathleen V. Gregg, Nicholas Hainen, Phillip Lampe, Cynthia Roepcke, Dingrando, Laurel, Kathleen V. Gregg, Nicholas Hainen, Phillip Lampe, Cynthia Roepcke,
and Cheryl Wistrom. Chemistry: Matter and Change. 1st ed. Vol. 1. Columbus: and Cheryl Wistrom. Chemistry: Matter and Change. 1st ed. Vol. 1. Columbus: Glencoe/ McGraw Hill, 2002. Glencoe/ McGraw Hill, 2002.
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The SIMPress Escape to End the Learning Module.
