The+Mole+(Period+D+H1)

__**Chapter 10: The Mole**__


 * Editor -** Jonathan Bailey

Period: D Chemistry I H1 Mr. Desjardins

This chapter is about measuring matter quantitatively, which is what moles allow you to do. By reading and understanding this chapter, you guys will be able to do things such as find the molar mass of different substances, as well as find the mass and volume relationships between those substances! I outlined which group everybody is in and what pages they are doing so if you don't remember just look at the list here! **__DON'T FORGET__** if you want Mole Bucks you can make a __video__ of yourself explaining a concept about the mole and how to do the math for it, **make sure to get it done** so we all get a good grade! Thanks guys!

Note: Feel free to insert your work underneath your name and page number that I put below, __please keep it neat!__ :)

__**Due Date = Thursday night, March 1st, 2012. Page closes at 6:00 PM!!!**__

__**Section 1: The Mole: A Measurement of Matter (Pages 287-296)**__

Pages 287-288 --- Marie Wachter

**Measuring Matter**
 * Chemistry is a quantitative science
 * how much/how many?
 * Ways to measure matter
 * count how much of something you have
 * count number of CDs
 * determine mass

__1 dozen apples__ __2.0 kg apples__ __1 dozen apples__ 12 apples 1 dozen apples .20 bushel apples
 * measure my volume
 * Some units indicate a specific amount
 * a pair means two
 * a dozen is 12
 * some things can be measured in several ways
 * Apples
 * by: 3 for $2.40
 * by weight: $1.28 per pound
 * by volume: $12.00 per bushel
 * Knowing how these relate, you can convert among units

This is a video showing how to convert moles to number of atoms. Sean filmed, Hanna narrated, Marie wrote. media type="file" key="IMG_0676.m4v"

Page290 --- Emily Crawford

What is a Mole?

 * Matter is composed of atoms, molecules, and ions.
 * Particles are very, very small.
 * A mole is the unit chemists use to group particles.
 * **Mole- 6.02 X 10^23 representative particles**
 * It is the SI unit for measuring the amount of a substance.
 * **Avogadro's Number-**6.02 X 10^23
 * Helped clarify the difference between atoms and molecules.
 * **Representative Particles- the species present in a substance (atoms, molecules, or formula units)**
 * For most elements it is the atom.
 * Ex) iron is made of iron atoms
 * Seven elements are diatomic molecules.


 * Ex) Molecular Compound Water= H20
 * Ex) Ionic Compound Calcium Chloride= CaCl2
 * A mole of any substance contains Avogadro's number of representative particles, or 6.02 X10^23 representative particles.

Pages 290-293 --- Hannah Mullen __Converting Number of Particles to Moles __ __Converting Moles to Number of Particles __
 * Conversion factor: 1 mol = 6.02 X 10^23
 * Used to determine moles
 * Moles = representative particles X 1 mol / (6.02 X 10^23 rep particles)
 * Determine # of atoms in 1 mole of compound
 * You must find the number of atoms in a representative particle to do this
 * Representative Particles = moles X (6.02 X10^23 rep particles) / 1 mol
 * Finds atoms in one mole of compound


 * Avogadro's number is huge. By his number the mass of the animal mole is 9.03 X 10^22

Pages 293-294 --- Ellen Mahoney **Mass of a Mole of an Element** The atomic mass of an element is expressed in atomic mass of units (amu) Refer to table 10.2: In the periodic table, atomic masses are not written as whole numbers
 * Atomic masses: relative values based on mass of most common isotope of carbon
 * atomic mass of average carbon atom (C) = 12.0 amu (12 times heavier than the average hydrogen atom, H, whose atomic mass is 1.0 amu)
 * 100 carbon atoms = 12 * 100 hydrogen atoms (this works for any of the same number of carbon atoms and hydrogen atoms)
 * this mass ratio is the same, no matter what the unit is (so 1.0 g of hydrogen has the same number of atoms as 12.0 g of carbon atoms)
 * this is because atomic masses are weighted average masses of the isotopes of each element

Page 294 --- Elena Berube

Molar Mass

 * The Atomic Mass of an element expressed in grams is the mass of a mole of the element.
 * **Molar Mass** is the mas of a mole of an element.
 * The molar mass of an element is usually very similar to the atomic mass of an element that you find on your periodic table.
 * When you compare the molar mass of 2 elements, they must contain the same number or atoms, which is 6.02 x 10^23 atoms of that element
 * The mole can be defined ass the amount of substance that contains as many representative particles as the number of atoms in 12.0 grams in Carbon-12. The same relationship applies to hydrogen: 1.0 g of hydrogen is equal to 1 mole of hydrogen atoms.

Pages 295-296 --- Sean Lydon

The Mass of a Mole of a Compound

 * To determine the mass of a mole of a mole of a compound, you must know the formula of the compound. For example, the formula of Carbon dioxide is CO 2.
 * To calculate the mass of a molecule, add the atomic masses of the atoms in the molecule. For example, the molar mass of CO 2 is equal to the mass of one atom of carbon plus the mass of two atoms of oxygen: (C = 1 x 12.01 amu) + (O = 2 x 16 amu) = 12.01 amu + 32 amu = 44.01 amu.
 * Then switch the amu unit with the grams unit to show that this number is the molar mass of the compound. Therefore, one mole of Carbon dioxide (CO 2 ) has a mass of 44.01 g.
 * This method can be applied to any compound, molecular or ionic.
 * Now let's try zinc nitrite. The formula for Zinc nitrite is Zn(NO 2 ) 2 . The equation to calculate the molar mass of this compound is (Zn = 1 x 65.39 amu) + (N = 2 x 14.01 amu) + (O = 4 x 16 amu) = 65.39 amu + 28.02 amu + 64 amu = 157.41 amu. One mole of Zinc nitrite [Zn(NO 2 ) 2 ] has mass of 157.41 g.

Here is a video that explains this method in greater detail and also provides some examples: media type="youtube" key="F9NkYSKJifs" height="358" width="516"

Elena and I made a quick video describing how to find the molar mass of a compound! media type="custom" key="12900666" __**Section 2: Mole Mass and Volume Relationships (Pages 297-303)**__

Page 297 --- Melanie Brondyk
 * The Mole-Mass Relationship
 * Molar mass of any substance is the mass in grams of one mole of that substance
 * ALL substances: elements, molecular compounds, and ionic compounds
 * Sometimes, molar mass may be unclear
 * Can avoid confusion by using the formula of the substance
 * Use the molar mass of an element or compound to convert between the mass of a substance and the moles of a substance
 * molar mass=1 mole
 * To find the mass in grams of a given number of moles:
 * Mass (grams)= number of moles * mass (grams)/ 1 mole

Pages 298-299 --- Korey Dufault Converting Moles to Mass and Converting Mass to Moles:


 * These photo's help you learn how convert moles to mass. Sample problem 10.6
 * First, you have to analyze the problem by listing the know and the unknown.
 * Second, calculate and solve for the unknown.
 * Third, evaluate and see if the answer makes sense.
 * There are also a couple practice problems at the bottom of the page.
 * In this sample problem above, you used a conversion factor based on the molar mass to convert moles to mass. 1 mol = molar mass.






 * These photo's help you learn about how to convert Mass to Moles. This is sample problem 10.6
 * You follow the same strategy by analyzing, calculating, and evaluating your work.
 * There are also a couple practice problems to give you more experience with doing these types of problems.

media type="youtube" key="T9Agc77ulTw" height="315" width="420"


 * This video will help you in converting Moles to Mass.

media type="youtube" key="WlGqMize7tI" height="315" width="420"


 * This video will you in converting Mass to Moles.

Page 300 --- Darby Barrett

The Mole-Volume Relationship
This video will explain the mole-volume relationship and Avogadro's hypothesis



Page 301 --- Lexie St. Jacques

media type="youtube" key="SXo13S_B9f4" width="425" height="350"


 * The molar volume is used to convert a known number of moles of gas to the volume of the gas at STP.**

Conversion factor: 22.4 L = 1 mol at STP
(Volume of gas = moles of gas x 22.4L / 1mole)


 * Example problem:**



Step 1



Step 2



Step 3



The opposite conversion (from volume of a gas at STP to the number of moles in a gas) uses the same factor.

Page 302 --- Sam Massoud


 * Calculating Molar Mass from Density **

**The density of a gas at STP and the molar volume at STP (22.4 L/mol) can be used to calculate the molar mass of the gas.**
 * STP is the conditions under which the volume of a gas is usually measured
 * Reminder: density = mass/volume

Molar mass = density at STP x molar volume at STP, or  ** __grams__ = __grams__ x __22.4 L__ ** ** mole L 1 mole ** <- here is an example of a molar mass by density problem

Page 303 --- Joe Jennings

//When Converting Mass to Moles://
 * In Summary:**

__Mass x (1.00mol/molar mass) = Mole__

// When Converting Moles to Mass: //

__Moles x molar mass/1.00mol) = Mass__

//When Converting Representative Particles to Moles://

__Representative Particles x (1.00mol/6.0210^23 particles) = Mole__

//When Converting Moles to Representative Particles://

__Mole x (6.02x10^23 particles/1.00mol) = Rep. Particles__

//When Converting Volume of Gas to Moles://

__STP x (1.00 mol/22.4L) = Moles__

//When Converting Moles to Volume of Gas//

__Moles x (22.4L/1.00mol) = STP__

__**Section 3: Percent Composition and Chemical Formulas (Pages 305-312)**__

Page 305 --- Nick Achin

The Percent Composition of a Compound

 * The relative amounts of elements in a compound are shown in the percent composition.
 * The percent composition is the percentage of each element in a compound.
 * The formula to find the percent of an element in a compound is:
 * %mass of element= mass of element/mass of compound x 100%**

Page 305-306 --- Lauren Berube

Analytical procedures are used to determine masses and calculate percent composition.
Math: Analyze: -List knowns and unknowns. Calculate: -Solve for unknowns. Evaluate: -Does the result make sense?

Example: 13.60g sample of compound of magnesium and oxygen is decomposed, 5.40g of oxygen is obtained. What is the percent composition of this compound?

Analyze: -Knowns mass of compound- 13.60 g mass of oxygen- 5.40g mass of mg- 13.60-5.40= 8.20g Mg -Unknowns percent of Mg percent of O percent mass= mass of element divided by mass of compound multiplied by 100%.

Calculate: Mg= __mass of mg__ x100 Mass of compound O= __mass of o__ x100 Mass of compound

Evaluate: 60.3+39.7=100 yes, it makes sense

Page 307 --- Kyle St. Pierre

Percent Composition from the Chemical Formula

- The percent composition of a compound can be calculated with just the chemical formula - This will give the percent composition
 * Multiply the subscript by the atomic mass on each atom
 * Add up all of these masses to get the molar mass
 * Divide the individual mass of an atom by the molar mass and multiply by 100%

Here is an example:


 * Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane.**

Analyze: List the knowns and unknowns

Knowns -Mass of C in 1 mol C3H8 = 36.0g -Mass of H in 1 mol C3H8 = 8.0g -Molar Mass of C3H8 = 44.0g/mol

Unknowns: Solve for the Unknowns - Percent C = ? %C - Percent H = ? %H

Calculate: Solve for the Unknowns

%C = mass of C / mass of propane X 100% = 36.0g / 44.0g X 100% = 81.8%

%H = mass of H / mass of propane X 100% = 8.0g / 44.0g X 100% = 18%

Evaluate: Does the Result Make Sense?

If the percent of the elements adds up to about 100% then the answer is correct. This answer does add up to 100% so it is correct

__This is a video of a teacher giving a lecture on how to solve a percent composition formula, so it should be helpful in gaining full comprehension of the topic:__

media type="youtube" key="uRVOac-7REU" height="315" width="420" Page 308 --- Brittany Chlebek

Percent Composition as a Conversion Factor
 * you can use percent composition to calculate the number of grams of any element in a specific mass of a compound
 * Finding Percent Composition:
 * multiply the mass of the compound by a conversion factor based on the percent composition of the element in the compound
 * Example: propane - you want to know how much carbon and hydrogen are contained
 * propane is 81.8% carbon and 18% hydrogen
 * in a 100-g sample of propane, you would have 81.8 g of carbon and 18 g of hydrogen
 * use the ratio 81.8 g C/100 g C3H8 to calculate the mass of carbon contained in 82.0 g of propane (C3H8)
 * 82.0 g C3H8 x (81.8 g C/100 g C3H8) = 67.1 g C
 * use the ration 18 g H/100 g C3H8 to calculate the mass of hydrogen
 * 82.0 g C3H8 x (18 g H/100 g C3H8) = 15 g H
 * the sum of the 2 masses equals 82 g, the sample size, to two significant figures

Pages 309-310 --- Maddie Harmon

Empirical Formula

 * Shows a basic ratio of elements
 * You can multiply the ratio by any factor to produce formulas for other compositions
 * In order to calculate the basic ratio of elements, you need the percent composition of the newly synthesized compound
 * The basic ratio is called the empirical formula.
 * It gives the **lowest** whole-number ratio of the atoms of elements in a compound
 * Shows the kinds and lowest relative count of atoms, molecules in atoms, or formula units of a compound
 * The empirical formula may or may not be the same as the Molecular formula
 * for example, The empirical formula for Hydrogen peroxide is HO.
 * The actual molecular formula for hydrogen peroxide is H202
 * The ratio for both is 1:1
 * **The empirical formula of a compound shows the smallest whole-number ratio of the atoms in the compound.**
 * Molecular formula tells the actual number of each atom
 * To find the empirical formula of a compound, you must first find the percent composition for each atom in the compound

1. First, you have to list the knowns and unknowns. 2. Then, you solve for the unknown.

3. Then, you evaluate your answer to see if it makes sese. Here are some practice problems:
 * The subscripts are whole numbers, adn the percent composition of this empirical formula equals the percents given in the original problem.

Here is a youtube video that can help you learn how to create the empirical formula: []

Pages 311-312 --- Kyle Mahoney

=Molecular Formulas=

//**The Molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula.**// Once you determine the empirical formula, you can determine its molecular formula, __But you must know its molar mass.__
 * EFM = Empirical Formula Mass
 * The molar mass represented by the empirical formula.

TO GET THE MULTIPLIER FOR CONVERSION

 * 1) Divide the experimentally determined mass by the empirical formula mass.
 * 2) Step 1 gives you the multiplier for converting it to the molecular formula

PROBLEM WITH EXPLANATION TO FIND MOLECULAR FORMULA
Ex. Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH4N. Step 1: Figure out what you know You don't know the molecular formula yet, but you will, I promise. Step 2: Solve First calculate the empirical formula mass. Then divide the molar mass by the empirical formula mass to obtain a whole number. To get the molecular formula, multiply the formula subscripts by this value.
 * the Empirical Formual is CH4N
 * Molar Mass 60.0g/mol


 * Empirical Formula || EFM || Molar mass/EFM || Molecular Formula ||
 * CH4N || 30.0 || 60.0/30.0 = 2 || C2H8N2 ||