Day 48 - Reactions and Balancing Quest, Introduction to Moles

Tuesday, November 13, 2012

By: Evan Ludig


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:

  • There MUST be some sort of proof that the student themselves did the baking, and not their parents.


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)download.jpg
  • 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.

Useful References:
  • Section 6.3 (pg 266-270), Section 6.4 (pg 271- 277), and Section 6.5 (pg 278-283)

  • Complete worksheet on Avagadro's Number, Moles and Molar Mass by Monday, November 19th, 2012
  • Review your notes daily!

Day 50 - Moles and Molar Mass (Thursday November 15th, 2012)
By Nilab Ahmaddi

  • Work period to complete worksheet on Avagadro's Number, Moles and Molar Mass
  • Another worksheet was given (Mixed Mole Practice Problems) to be completed starting from question 9 and on.

Overview of Molar Mass
  • Molar Mass: mass of 1 mole of a substance
  • Symbol used is MM
  • The unit is g/mol

* the 2 formulas for calculating the # of moles *
converting number of particles to moles
converting mass into moles
n= # of particles / NA
- n= # of moles
- NA= Avogadro's #
m= mass in grams
MM= molar mass in g/mol
- Converting number of particles to moles

- The Mole and Avogadro's Number

  • Complete Avagardro’s Number, Moles and Molar Mass for Monday November 19th, 2012
  • Complete Mixed Mole Practice Problems (Q#9-19) for Monday November 19th, 2012
  • "Everything must go through moles!!!” handout is due Monday November 19th, 2012

  • COOKIES ISU is due Monday November 26th, 2012

Day 51 - Mole Calculations

By: Evan Ludig


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.


Students were to complete the second sheet given to them, which included problems on mass, number of moles, and number of particles


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: Sucrose-C12H22O11
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 #1-2 Practice questions 4-12 bottom of page
  • Page 280 questions #1-4
  • Page 283 questions #4-6

  • COOKIES ISU is due Monday November 26th, 2012

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

external image Percent+Composition+Formula.bmp

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:
Finish Percentage Composition exercise #1-9
  • Quiz tomorrow on moles
  • COOKIES ISU is due Monday November 26th, 2012

Day 54- November 22nd, 2012

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?

Mass or %
Molar mass (MM)
n=Mass or %
Molar mass
Divide by smallest from column 4
Reduce to smallest whole number ratio

The simplest formula is Na2CO3• 10H2O

Do empirical formula worksheet
  • 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

  • Did a lab on determining the Formula of a Hydrate

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.

external image crucible.apparatus.gif

Extra videos to help prepare for Monday’s quest!

  • Moles, mass and molar mass calculation

  • Percent composition and Empirical formula

  • STUDY FOR QUEST ON MONDAY NOVEMBER 26th, 2012 à[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, 2012 à 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


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:
    1. Write a balanced chemical equation for the reaction.
    2. Identify the known and unknown materials in the problem.
    3. Calculate the number of moles of the given or known material.
    4. Use the molar ratio (see example problem below) from the balanced chemical equation to determine the number of mole from the unknown.
    5. 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


122.54 g/mol
32.00 g/mol


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.

3 / 2

x / 0.1000 = 4.800g

x = 0.1500


Here is a helpful video on stoichiometry.


For homework we were to complete some questions on the reverse side of the sheet given.


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


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.


Steps to finding the limiting factor (LF):
  1. Write a balanced chemical equation.
  2. Create a table with all known quantities (e.g. mass, molar mass moles)
  3. Divide the number of moles of each known species by its coefficient.
  4. 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 à 2NH3

Step 2)

? 2.43g

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.


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


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;
  1. competing reactions- a reaction that occurs at the same time as the principal reaction.
  2. lab technique- spilling a product or leaving product behind on glassware.
  3. reaction condition- some reactions may need time to react, others need different conditions (temperature).
  4. reactant purity- assuming that all reactants are pure.


Steps to finding the percent yield:
  1. Write a balanced chemical equation.
  2. Create a table with all known values (e.g. mass, molar mass, moles).
  3. Solve for mass of wanted species (theoretical yield)
  4. 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)
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%.


p. H5 Practice Questions #1-13

Day 59 – November 30th, 2012 (Work Period)

By: Nilab Ahmaddi


- 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.


Things covered in this unit:

  • Significant digits à 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 chemicals à

- 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),

  • The moleà there are 6.02 x10 23 particles in one mole, molar mass is calculated from the periodic table

  • Percentage composition à Element mass/ compound mass x 100%

  • Simplest and molecular formulae à determined by the percent composition, grams of reactants or moles

  • The factor label method à 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:
  1. STUDY FOR Unit 4 Test TUESDAY DECEMBER 4th, 2012 !
  2. Complete the Unit 4 Review questions
  3. Complete observation table for MONDAY DECEMBER 3RD, 2012
  4. Return trip forms!


Day 62 – Tuesday, December 4, 2012


By: Amentha Pusparajah


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

external image logo.gif

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:


    • if you need further assistance, please go see Ms.Wilson during clinic.


  • Tomorrow we will be starting a new unit solutions
  • We will be making observations on our lab conducted yesterday and will be measuring the mass of the filter paper
  • The good copy of the lab will be due Monday
    • observation table
    • analyze and evaluate (a-h) and calculate percent yeild
    • conclusion (source of error/improvements
    • rough copy
    • reference (APA Format)
      • Textbook
      • Lab partner
      • Ms. Wilson

  • Dont forget to being permission forms and money for our field trip on December 18th

Homework: n/a