Day 48  Reactions and Balancing Quest, Introduction to Moles
Tuesday, November 13, 2012
By: Evan Ludig
Summary:
Today in class we had a quest on reactions and balancing equations for the first half, followed by an introduction to moles and received the cookie ISU.
Pre  Class Requirements
Students required their answer sheet from the questions yesterday to be handed in for class today.
Class Overview
Had a quest comprised of 6 questions on chemical reactions and balancing for approximately the first 30 minutes of class.
Received two handouts, the first being an introduction to moles and Avagadro's number (6.02 x 1023 ).
Used pea demonstration to show the true size of Avagadro's number, which with snow peas covers the nearest 250 Earth sized planets up to 1 metre!
The second sheet was an instruction sheet for the next ISU, bake a batch of cookies following the instructions
Measurements must be calculated from moles to grams, litres etc.
ISU is due on Monday, November 26, at least one cookie must be brought in with pre  lab questions (math work).
Additional Reminders:
DO NOT BRING COOKIES IN TUPPERWARE CONTAINERS! A plastic bag will suffice.
There MUST be some sort of proof that the student themselves did the baking, and not their parents.
Moles
Avagadro's Number, Moles, And Molar Mass November 14th, 2012
By: Ajaykumar Shanmugaraj Summary: Today's lesson was based on the following concepts:
Molar Mass (MM)
Calculations Using the Mole
o Formula 1: Converting number of particles to moles
o Formula 2: Converting mass into moles
Avagadro's Number (NA)
Overview of Today's Lesson:
Students were taught on how to calculate Molar Mass with a given element or compound
Students were also taught conversions using the mole
Mole (n) is the units used in chemistry
Avagadro's Number:6.02 x 1023 particles/mol
o This constant value represents the number of entities in one mole
1 Mole = 6.02 x 1023
Molar Mass:
Definition: Is the mass of 1 mole of a pure substance (either an element or compound)
Symbol: MM
Unit:grams/mole (g/mol)
The numerical value on the Periodic Table for atomic mass is the same for molar mass. However, instead of being measured in atomic mass units, it is measured in g/mol
How to Calculate Molar Mass for Compounds:
1. Write the chemical formula of the compound.
2. Multiply the atomic mass for each element by its subscript in the chemical formula.
3. Add all the masses. Final answer should be in 2 decimal places and measured in g/mol.
Mole Calculations: 1.Converting # of particles to moles:
Formula: Moles= #of Particles/ Avagadro's #
n= of particles/ NA
Round final answer according to correct number of significant digits.
2.[[#Convert]] mass into moles Formula:Moles = Mass/ Molecular Mass ;
n = m/ MM
Round final answer according to correct number of significant digits.
Complete Avagardro’s Number, Moles and Molar Mass for Monday November 19th, 2012
Complete Mixed Mole Practice Problems (Q#919) for Monday November 19th, 2012
"Everything must go through moles!!!” handout is due Monday November 19th, 2012
Reminder:
COOKIES ISU is due Monday November 26th, 2012
Day 51  Mole Calculations
By: Evan Ludig
Summary:
Today in class we received 2 worksheets, one being a fun worksheet were the answer upside down spelt a word, the other practice problems for finding moles of atoms, mass, molar mass, etc.
Class Overview:
Completed a worksheet where for the questions the answer was a word upside down, to fill in sentences on the back of the page
Practice Problem:
1.a) i. The mass of 100 of Arsenic = 7,500g
ii. Add the mass of 1 mol of Cesium to i. = 133 + 7,500 = 7,633
iii. Add 101 to ii. = 7,633 + 101 = 7,734
Flipped upside down, iii. is HELL
After completion of previous worksheet, students were given another worksheet with practice problems to be completed.
Example Problem:
2.a) How many moles of atoms are there in 2 moles of H2O?
i. We can tell that there are only 3 atoms in the molecule.
ii. To normally find the number of molecules, we would use # of moles x Avagadro's number.
iii. 2 x 6.02 x 1023 = 1.20 x 1024 , now multiply this number by the number of atoms
iv. 3 x 1.20 x 1024 = 3.60 x 1024 is the number of atoms in an H2O molecule.
Homework:
Students were to complete the second sheet given to them, which included problems on mass, number of moles, and number of particles
Reminders:
COOKIE ISU DUE IN 1 WEEK! Make sure to bring cookies in any way other than Tupperware!
Day 52 – How many molecules: In class Assignment (Tuesday November 20th, 2012) By Andrew Samuel R Assignment Summary:
3 Stations set throughout the class
Individually go to each station and follow procedure set by you
Goal is to find how many molecules are in each substance
The three questions: How many molecuels of water H2O
How many molecuels in 1 cube of sugar C12H22O11
How many molecuels on the sheet of paper written in chalk
Overview of Assignment
First estimate the number of molecules there would be for each test
Write down your own procedure
Go to each station and follow the procedure
Once got data needed go back to desk and use formula to find the number of molecules
Hand it in once done with procedure
Step 1: Needed to find the mass by using the weighing machine. There was one at each station.
Step 2: Needed to find molar mass. Either using the formula n =m/MM OR
Simply add the atomic mass and multiplying by its subscript.
Ex: SucroseC12H22O11
MM= 12C + 22H +11O
= 12(12.01) +22(1.01) +11(16.0)
= 342.34 g/mol
Step 3: Now you have both the molar mass and mass. Use the formula n=m/MM.Take the mass and divide by molar mass, this gives you moles.
Step 4: Use the formula n = #of Particles/NA Take the number of moles and multiply it with Avogadro’s number. Avogadro’s number = 6.02 × 1023
Extra Practice:
Page 277 questions #12 Practice questions 412 bottom of page
Page 280 questions #14
Page 283 questions #46
Reminder:
COOKIES ISU is due Monday November 26th, 2012
POP QUIZ ON MOLES THIS WEEK!!
Day 53 – Percentage Composition (Wednesday November 21st, 2012) By Andrew Samuel R Summary: Today's lesson was on Percentage Composition
Percentage composition is the percent by mass of each element in a compound. This value gives the fraction of each element in a compound and can be calculated by 2 ways. 1.Using Experimental Data 2.Using the Chemical Formula of a Compound
Overview of Today's Lesson: Students were taught on how to find percent composition for each method Experimental Data: In this method you need to know the total mass of the compound and the mass of each individual element. Use the formula
Example:A 45.9 g sample of Copper, Cu, combined with oxygen to give 51.69 g of an oxide compound. What is the percent composition of each element?
%Cu= mass Cu/51.69g ×100% %O=%100 – percent of Cu %Cu = 45.9g/51.69g×100% %O =%100%88.8 %Cu=88.8% %O= 11.2%
Chemical formula:
How to find the percent composition of a compound: 1. Write a correct formula for the compound 2. Find the molar mass of the compound 3. Divide the total atomic mass of EACH ELEMENT by the molar mass 4. Multiply by 100 to convert your results to a percent
5. Since you have no significant figures to go by, express your answer to TWO decimal places with the % sign.
In this method the molar mass of the compound must be calculated and the total mass of each element. Use the formula % Composition of an element = #of atoms of an element × atomic mass/ Molar Mass of compound ×100%
Example: Calculate the percent by weight of sodium (Na) and chlorine (Cl) in sodium chloride (NaCl)
Calculate the molecular mass (MM):MM = 22.99 + 35.45 = 58.44
Calculate the total mass of Na:1 Na, mass = 22.99
Calculate the percent by weight of Na in NaCl:%Na
(mass Na ÷ MM) x 100
(22.99 ÷ 58.44) x 100 = 39.34%
Calculate the total mass of Cl:1 Cl , mass = 35.45
Calculate the percent by weight of Cl in NaCl:%Cl
By: Andrew Samuel Summary: Today's lesson was on Empirical formula
Overview of Today's Lesson: Students were taught on how to use the Empirical formula. What it is: Is the simplest ratio of elements in a compound. It uses the smallest possible whole number ratio of atoms in a formula of a compound. With the percent composition, an empirical formula can be calculated. Ionic compounds are always expressed in the simplest formulas. While covalent compounds can be both molecular formulas or simplest.
EX: Hydrogen peroxide True formula: H2O2 EmpiricalFormula: HO
How to find the empirical formula of a compound: % mass Mass to mole Divide by small Multiply till whole
It is always good to use this chart format when finding empirical formula
EX: A 5.72g sample of washing soda (Na2CO3 • xH2O) is heated to give 2.12g of anhydrous Na2CO3. What is the simplest formula of the hydrated salt?
Species
Mass or %
Molar mass (MM)
n=Mass or % Molar mass
Divide by smallest from column 4
Reduce to smallest whole number ratio
Na2CO3
2.12g
105.99
2.12/105.99=0.02
1
1
H2O
3.6g
18.02
3.6/18.02=0.1998
9.99
10
The simplest formula is Na2CO3• 10H2O
Homework: Do empirical formula worksheet Reminder:
Quest Monday on moles, percent composition, and Empirical Formula
COOKIES ISU is due Monday November 26th, 2012
Day 55 November 23nd, 2012
By: Nilab Ahmaddi
Summary
Did a lab on determining the Formula of a Hydrate
Overview
Today a lab was done throughout most of the period. We worked together in pairs to observe and remove the water from a hydrated form of magnesium sulfate which will allow us to determine the percentage composition of water. Within this lab, we placed a clay triangle on an iron ring and placed the lit Bunsen burner underneath it. The crucible was heated for about 5 minutes and was left to cool. At this time the mass of the empty crucible was taken and about 3 grams of hydrate was then added to the crucible which was left to heat for about 10 minutes and cooled for about 2 minutes. Then the mass was taken again, reheated for about 2 minutes and left to cool before taking the mass once more.
NOTE* This lab will not be done as a formula lab; instead the calculations will be finished in class on Monday November 26th 2012 once you have completed the quest.
STUDY FOR QUEST ON MONDAY NOVEMBER 26th, 2012à[Moles, percent composition and empirical formula], make sure to complete all sheets that have been given from this section for practice.
COOKIE ISU DUE MONDAY NOVEMBER 26th, 2012à Complete and show all work for the pre lab questions that is to be handed in with your cookie. Also, please provide evidence that you have baked these cookies (note from a parent or a picture). Make sure to bring only one cookie in a zip lock bag to be handed in.
Day 57  Stoichiometry Mass to Moles
Tuesday, November 27, 2012
By: Evan Ludig
Summary:
Today in class we had an introductory lesson to stoichiometry, which is the quantitative relationships between products and reactants (A.K.A basically finding an unknown value between the products and reactants of a chemical
reaction, such as the amount of chemical A when chemical B decomposes). We also received 3 worksheets on the subject.
Class Overview:
Steps to solving a stoichiometry problem:
Write a balanced chemical equation for the reaction.
Identify the known and unknown materials in the problem.
Calculate the number of moles of the given or known material.
Use the molar ratio (see example problem below) from the balanced chemical equation to determine the number of mole from the unknown.
Convert the number of moles of the unknown material to whatever is asked for in the question.
Example Problem:
Calculate the mass of oxygen gas produced when 12.26g of potassium chlorate decomposes into potassium chloride and oxygen
Balanced equation:
2 KClO3 > 2KCl + O2
Table:
KClO3
KCl
O2
Molar
Ratio
2
2
3
Mass
12.26g

?
Molar
Mass
122.54 g/mol

32.00 g/mol
Moles
0.1000n

Calculations:
x = number of moles of oxygen Oxygen's mass = moles x molar mass Therefore, the mass of the oxygen gas produced
= 0.1500 x 32.00 is 4.800g.
O2
For homework we were to complete some questions on the reverse side of the sheet given.
Reminders:
THIRD ARTICLE IS DUE ON FRIDAY! Do not forget to bring it in!
Day 58 Calculations Involving Limiting Factors
Wednesday, November 28, 2012
by: Emilija Milenkovski
SUMMARY:
Today in class, we learned how to calculate the limiting factor in a chemical equation. The limiting factor determines the amount of each product that is needed if there is a limit on the amount of some of the species. We also learned how to find the mass of the excess reagent left at the end of the reaction.
CLASS OVERVIEW:
Steps to finding the limiting factor (LF):
Write a balanced chemical equation.
Create a table with all known quantities (e.g. mass, molar mass moles)
Divide the number of moles of each known species by its coefficient.
The smaller value is the limiting factor.
Example: When 1.50g of hydrogen reacts with 2.00g of nitrogen, how much ammonia is formed? Find the mass of ammonia, and the excess reagent left over after the reaction.
Step 1) 2H2 + N2 à2NH3
nN / MRN = 0.0714mol / 1 = 0.0714
the smallest amount of moles might not always the limiting factor
Step 4) Therefore the limiting factor is nitrogen.
To find the mass of ammonia, use the limiting factor (N) to determine the number of moles in NH3
Cross multiply: x = 2(0.0714) = 0.143mol Convert to mass: m = nMM = 0.143mol (17.04g/mol) = 2.43g
To find the excess amount of hydrogen, use the limiting factor (N) in a ratio equation, then useformula(below)
Cross multiply: x = 2(0.0714)
= 0.143mol
Therefore 0.143 moles are used in the reaction.
excess = moles avaliable  moles used
= 0.743mol  0.143mol
= 0.600mol
Convert to mass: m = nMM
= 0.600mol (2.02g/mol)
= 1.21g
Therefore, there is 1.21g of hydrogen leftover.
Homework:
Complete the questions on the back of the sheet given (Calculating Involving Limiting Factors)
Day 59 Percent Yield
Thursday, November. 29, 2012
By: Emilija Milenkovski
SUMMARY:
Today, we learned percent yield, which is the percent of product that was actually formed. A simple example would be calculating the percent you actually got on a test. Factors that lead to percent yield are;
competing reactions a reaction that occurs at the same time as the principal reaction.
lab technique spilling a product or leaving product behind on glassware.
reaction condition some reactions may need time to react, others need different conditions (temperature).
reactant purity assuming that all reactants are pure.
CLASS OVERVIEW:
Steps to finding the percent yield:
Write a balanced chemical equation.
Create a table with all known values (e.g. mass, molar mass, moles).
Solve for mass of wanted species (theoretical yield)
Divide actual yield (usually given in the question) by the theoretical yield, and multiply by 100
Example: If 5.00g of C2H5OH is recovered after 16.0g of C6H12O6, what is the percentage yield of the reaction?
Step 1) C6H12O6 à 2C2H5OH + 2CO2
Step 2)
molar
ratio
1
2
2
mass
16.0g
5.00g
/
molar
mass
180.18g/mol
46.08g/mol
/
moles
0.0888mol
0.178mol
/
Calculations:
Step 3) Find moles of C2H5OH and convert to grams.
Cross multiply: x = 2(0.0888)
= 0.178mol
Convert to mass: m = nMM
= 0.178mol (46.08g/mol)
= 8.20g
Step 4) Therefore, theoretical yield is 8.20g.
To find the percent yield, useformula(below) % yield = actual yield / theoretical yield x 100%
= 5.00 / 8.20 x 100%
= 61.0%
Therefore the percent yield is 61.0%.
Homework:
p. H5 Practice Questions #113
Day 59 – November 30th, 2012 (Work Period)
By: Nilab Ahmaddi
Summary:
 Today in class, we worked in groups to complete different worksheets (Unit 4 Review or the challenge problems) in order to prepare for Tuesday’s Test.
Overview:
Things covered in this unit:
Significant digitsà All digits to right of the first number are significant, in scientific notation all digits are significant , for multiplication/division: # digits= fewest and the average atomic mass is equal to the sum of individual isotope masses multiplied by their percentage
Balancing chemicalsà
 Balancing equations by inspection  Types of reaction: Decomposition (AB à A +B), Synthesis (A+B àAB), Single Displacement (AX+ YàYX +A), Double Displacement (AB+XY à AY + XB),
Combustion(complete/incomplete)
The moleà there are 6.02 x10 23 particles in one mole, molar mass is calculated from the periodic table
Percentage compositionà Element mass/ compound mass x 100%
Simplest and molecular formulaeà determined by the percent composition, grams of reactants or moles
The factor label methodà creating conversion factors. How to use the factor label method
Stoichiometryà grams x à moles x à moles y à grams y
Limiting reagentsà actual/ideal chart for limiting reagents. The limiting reagent is the given quantity
Percentage yields à percentage yield = actual/theoretical x 100%, the actual yield will be given, theoretical must be calculates
Homework/ Reminders:
STUDY FOR Unit 4 TestTUESDAY DECEMBER 4th, 2012 !
Complete the Unit 4 Review questions
Complete observation table for MONDAY DECEMBER 3RD, 2012
Return trip forms!
_
Day 62 – Tuesday, December 4, 2012
UNIT TEST CHEMICAL REACTIONS, MOLES, AND STOICHIOMETRY
By: Amentha Pusparajah
Overview:
Today we had our unit test on Chemical Reactions, Moles and Stoichiometry.
There was no lesson today because the entire period was taken up for our unit test that covered:
Types of Chemical Reactions
Synthesis
Double displacement
Single displacement
Decomposition
Combustion
Exothermic & Endothermic
Balancing Chemical Reactions
Moles
Molar mass
Percent Composition
Empirical and Molecular Formula
Limiting Factor
Percent Yield
Further Notes:
In a few days this page will be updated with the answers to the test in order to give you guys an opportunity to correct your mistakes.
If you need further assistance and practice for this unit, the following links will help you out:
Day 48  Reactions and Balancing Quest, Introduction to Moles
Tuesday, November 13, 2012
By: Evan Ludig
Summary:
Today in class we had a quest on reactions and balancing equations for the first half, followed by an introduction to moles and received the cookie ISU.
Pre  Class Requirements
Class Overview
Additional Reminders:
Moles
Avagadro's Number, Moles, And Molar Mass
November 14th, 2012
By: Ajaykumar Shanmugaraj
Summary: Today's lesson was based on the following concepts:
 Molar Mass (MM)
 Calculations Using the Mole
 o Formula 1: Converting number of particles to moles
 o Formula 2: Converting mass into moles
 Avagadro's Number (NA)
Overview of Today's Lesson: Students were taught on how to calculate Molar Mass with a given element or compound
 Students were also taught conversions using the mole
 Mole (n) is the units used in chemistry
 Avagadro's Number:6.02 x 1023 particles/mol
 o This constant value represents the number of entities in one mole
 1 Mole = 6.02 x 1023
Molar Mass: The numerical value on the Periodic Table for atomic mass is the same for molar mass. However, instead of being measured in atomic mass units, it is measured in g/mol
 How to Calculate Molar Mass for Compounds:
 1. Write the chemical formula of the compound.
 2. Multiply the atomic mass for each element by its subscript in the chemical formula.
 3. Add all the masses. Final answer should be in 2 decimal places and measured in g/mol.
Mole Calculations:1. Converting # of particles to moles:
Formula: Moles= #of Particles/ Avagadro's #
n= of particles/ NA
Round final answer according to correct number of significant digits.
2. [[#Convert]] mass into moles
Formula:Moles = Mass/ Molecular Mass ;
n = m/ MM
Round final answer according to correct number of significant digits.
Useful References:
Websites:
 http://www.chemteam.info/Mole/AvogadroNumberCalcsII.html
 http://herh.ccrsb.ca/staff/farrelll/molesFAQ.htm
 http://www.chem.memphis.edu/bridson/FundChem/T11a1100.htm
Quizes: http://www.sophia.org/convertingfromparticlestomolesconcept
 http://www.sophia.org/convertingfrommolestoparticlesconcept
 http://www.sophia.org/convertingfromgramstomolesconcept
 http://www.sophia.org/convertingfrommolestogramsconcept
Videos: http://www.youtube.com/watch?v=xc5Bh6o_gKk
 http://www.youtube.com/watch?v=ePyIYsszQjg&feature=relmfu
 http://www.youtube.com/watch?v=AsqEkF7hcII
Textbook:Homework:
Day 50  Moles and Molar Mass (Thursday November 15th, 2012)
By Nilab Ahmaddi
Summary:
Overview of Molar Mass
* the 2 formulas for calculating the # of moles *
 n= # of moles
 NA= Avogadro's #
m= mass in grams
MM= molar mass in g/mol
 Converting number of particles to moles
http://www.youtube.com/watch?v=tBbCX6dQZPo
 The Mole and Avogadro's Number
http://www.youtube.com/watch?v=AsqEkF7hcII
Homework:
Reminder:
Day 51  Mole Calculations
By: Evan Ludig
Summary:
Today in class we received 2 worksheets, one being a fun worksheet were the answer upside down spelt a word, the other practice problems for finding moles of atoms, mass, molar mass, etc.
Class Overview:
Homework:
Students were to complete the second sheet given to them, which included problems on mass, number of moles, and number of particles
Reminders:
COOKIE ISU DUE IN 1 WEEK! Make sure to bring cookies in any way other than Tupperware!
Day 52 – How many molecules: In class Assignment (Tuesday November 20th, 2012)
By Andrew Samuel R
Assignment Summary:
The three questions: How many molecuels of water H2O
How many molecuels in 1 cube of sugar C12H22O11
How many molecuels on the sheet of paper written in chalk
Overview of Assignment
Step 1: Needed to find the mass by using the weighing machine. There was one at each station.
Step 2: Needed to find molar mass. Either using the formula n =m/MM OR
Simply add the atomic mass and multiplying by its subscript.
Ex: SucroseC12H22O11
MM= 12C + 22H +11O
= 12(12.01) +22(1.01) +11(16.0)
= 342.34 g/mol
Step 3: Now you have both the molar mass and mass. Use the formula n=m/MM.Take the mass and divide by molar mass, this gives you moles.
Step 4: Use the formula n = #of Particles/NA Take the number of moles and multiply it with Avogadro’s number. Avogadro’s number = 6.02 × 1023
Extra Practice:
Reminder:
Day 53 – Percentage Composition (Wednesday November 21st, 2012)
By Andrew Samuel R
Summary: Today's lesson was on Percentage Composition
Percentage composition is the percent by mass of each element in a compound. This value gives the fraction of each element in a compound and can be calculated by 2 ways.
1.Using Experimental Data
2.Using the Chemical Formula of a Compound
Overview of Today's Lesson: Students were taught on how to find percent composition for each method
Experimental Data:
In this method you need to know the total mass of the compound and the mass of each individual element. Use the formula
Example:A 45.9 g sample of Copper, Cu, combined with oxygen to give 51.69 g of an oxide compound. What is the percent composition of each element?
%Cu= mass Cu/51.69g ×100% %O=%100 – percent of Cu
%Cu = 45.9g/51.69g×100% %O =%100%88.8
%Cu=88.8% %O= 11.2%
Chemical formula:
How to find the percent composition of a compound:1. Write a correct formula for the compound
2. Find the molar mass of the compound
3. Divide the total atomic mass of EACH ELEMENT by the molar mass
4. Multiply by 100 to convert your results to a percent
5. Since you have no significant figures to go by, express your answer to TWO decimal places with the % sign.
In this method the molar mass of the compound must be calculated and the total mass of each element. Use the formula
% Composition of an element = #of atoms of an element × atomic mass/ Molar Mass of compound ×100%
Example: Calculate the percent by weight of sodium (Na) and chlorine (Cl) in sodium chloride (NaCl)
Calculate the molecular mass (MM):MM = 22.99 + 35.45 = 58.44
Calculate the total mass of Na:1 Na, mass = 22.99
Calculate the percent by weight of Na in NaCl:%Na
(mass Na ÷ MM) x 100
(22.99 ÷ 58.44) x 100 = 39.34%Calculate the total mass of Cl:1 Cl , mass = 35.45
Calculate the percent by weight of Cl in NaCl:%Cl
(mass Cl ÷ MM) x 100
(35.45 ÷ 58.44) x 100 = 60.66%%Na + %Cl = 100, 39.34 + 60.66 = 100.
Useful References:
Websites:
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/percentcomp.htm
http://www.ausetute.com.au/percentc.html
http://www.kentchemistry.com/links/bonding/percentcomp.htm
Homework:
Finish Percentage Composition exercise #19
Reminder:
Day 54 November 22nd, 2012
By: Andrew SamuelSummary: Today's lesson was on Empirical formula
Overview of Today's Lesson:
Students were taught on how to use the Empirical formula.
What it is:
Is the simplest ratio of elements in a compound. It uses the smallest possible whole number ratio of atoms in a formula of a compound. With the percent composition, an empirical formula can be calculated. Ionic compounds are always expressed in the simplest formulas. While covalent compounds can be both molecular formulas or simplest.
EX: Hydrogen peroxide True formula: H2O2 EmpiricalFormula: HO
How to find the empirical formula of a compound:
% mass
Mass to mole
Divide by small
Multiply till whole
It is always good to use this chart format when finding empirical formula
EX: A 5.72g sample of washing soda (Na2CO3 • xH2O) is heated to give 2.12g of anhydrous Na2CO3. What is the simplest formula of the hydrated salt?
Molar mass
The simplest formula is Na2CO3• 10H2O
Homework:
Do empirical formula worksheet
Reminder:
Day 55 November 23nd, 2012
By: Nilab AhmaddiSummary
Overview
Today a lab was done throughout most of the period. We worked together in pairs to observe and remove the water from a hydrated form of magnesium sulfate which will allow us to determine the percentage composition of water. Within this lab, we placed a clay triangle on an iron ring and placed the lit Bunsen burner underneath it. The crucible was heated for about 5 minutes and was left to cool. At this time the mass of the empty crucible was taken and about 3 grams of hydrate was then added to the crucible which was left to heat for about 10 minutes and cooled for about 2 minutes. Then the mass was taken again, reheated for about 2 minutes and left to cool before taking the mass once more.
NOTE* This lab will not be done as a formula lab; instead the calculations will be finished in class on Monday November 26th 2012 once you have completed the quest.
Videos
Extra videos to help prepare for Monday’s quest!
http://www.youtube.com/watch?v=QcEyl4sGAc
http://www.youtube.com/watch?v=VgR9OV9idQs
Homework
Day 57  Stoichiometry Mass to Moles
Tuesday, November 27, 2012
By: Evan LudigSummary:
Today in class we had an introductory lesson to stoichiometry, which is the quantitative relationships between products and reactants (A.K.A basically finding an unknown value between the products and reactants of a chemical
reaction, such as the amount of chemical A when chemical B decomposes). We also received 3 worksheets on the subject.
Class Overview:
Balanced equation:
2 KClO3 > 2KCl + O2
Table:
Ratio
Mass
x = number of moles of oxygen Oxygen's mass = moles x molar mass Therefore, the mass of the oxygen gas produced
= 0.1500 x 32.00 is 4.800g.
O2
3 / 2
x / 0.1000 = 4.800gx = 0.1500
Video:
Here is a helpful video on stoichiometry.
Homework:
For homework we were to complete some questions on the reverse side of the sheet given.
Reminders:
THIRD ARTICLE IS DUE ON FRIDAY! Do not forget to bring it in!
Day 58 Calculations Involving Limiting Factors
Wednesday, November 28, 2012
by: Emilija Milenkovski
SUMMARY:
Today in class, we learned how to calculate the limiting factor in a chemical equation. The limiting factor determines the amount of each product that is needed if there is a limit on the amount of some of the species. We also learned how to find the mass of the excess reagent left at the end of the reaction.
CLASS OVERVIEW:
Steps to finding the limiting factor (LF):
Example: When 1.50g of hydrogen reacts with 2.00g of nitrogen, how much ammonia is formed? Find the mass of ammonia, and the excess reagent left over after the reaction.
Step 1) 2H2 + N2 à 2NH3
Step 2)
ratio
mass
factor
Step 3) moles/ coefficient = limiting
nH / MRH = 0.743mol / 2 = 0.372
nN / MRN = 0.0714mol / 1 = 0.0714
the smallest amount of moles might not always the limiting factor
Step 4) Therefore the limiting factor is nitrogen.
To find the mass of ammonia, use the limiting factor (N) to determine the number of moles in NH3
Cross multiply: x = 2(0.0714)
= 0.143mol
Convert to mass: m = nMM
= 0.143mol (17.04g/mol)
= 2.43g
To find the excess amount of hydrogen, use the limiting factor (N) in a ratio equation, then use formula (below)
Cross multiply: x = 2(0.0714)
= 0.143mol
Therefore 0.143 moles are used in the reaction.
excess = moles avaliable  moles used
= 0.743mol  0.143mol
= 0.600mol
Convert to mass: m = nMM
= 0.600mol (2.02g/mol)
= 1.21g
Therefore, there is 1.21g of hydrogen leftover.
Homework:
Complete the questions on the back of the sheet given (Calculating Involving Limiting Factors)
Day 59 Percent Yield
Thursday, November. 29, 2012
By: Emilija Milenkovski
SUMMARY:
Today, we learned percent yield, which is the percent of product that was actually formed. A simple example would be calculating the percent you actually got on a test. Factors that lead to percent yield are;
CLASS OVERVIEW:
Steps to finding the percent yield:
Example: If 5.00g of C2H5OH is recovered after 16.0g of C6H12O6, what is the percentage yield of the reaction?
Step 1) C6H12O6 à 2C2H5OH + 2CO2
Step 2)
ratio
mass
Step 3) Find moles of C2H5OH and convert to grams.
Cross multiply: x = 2(0.0888)
= 0.178mol
Convert to mass: m = nMM
= 0.178mol (46.08g/mol)
= 8.20g
Step 4) Therefore, theoretical yield is 8.20g.
To find the percent yield, use formula (below)
% yield = actual yield / theoretical yield x 100%
= 5.00 / 8.20 x 100%
= 61.0%
Therefore the percent yield is 61.0%.
Homework:
p. H5 Practice Questions #113
Day 59 – November 30th, 2012 (Work Period)
By: Nilab AhmaddiSummary:
 Today in class, we worked in groups to complete different worksheets (Unit 4 Review or the challenge problems) in order to prepare for Tuesday’s Test.
Overview:
Things covered in this unit:
 Balancing equations by inspection
 Types of reaction: Decomposition (AB à A +B), Synthesis (A+B àAB), Single Displacement (AX+ YàYX +A), Double Displacement (AB+XY à AY + XB),
Combustion(complete/incomplete)
Homework/ Reminders:
_
Day 62 – Tuesday, December 4, 2012
UNIT TEST CHEMICAL REACTIONS, MOLES, AND STOICHIOMETRYBy: Amentha Pusparajah
Overview:
Today we had our unit test on Chemical Reactions, Moles and Stoichiometry.There was no lesson today because the entire period was taken up for our unit test that covered:
Further Notes:
In a few days this page will be updated with the answers to the test in order to give you guys an opportunity to correct your mistakes.If you need further assistance and practice for this unit, the following links will help you out:
ü http://www.chemactive.com/examview/grade_11/stoichiometry.htm
ü http://www.grade11.com/content/view/2/40/
ü http://www.onlinemathlearning.com/stoichiometryproblems.html
Reminders:
Homework: n/a