Chapter+10



= Chapter 10 - Chemical Quanti﻿ties﻿ =

= The Mole Unit! =



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This page has been constructed on the tenth chapter of Prentice Hall’s //Chemistry// textbook. This chapter focuses on chemical quantities and the mole. This chapter specifically explains methods of measurement, and introduces the mole. It explains Avogadro’s number and hypothesis, along with molar mass, molar volume and their relationship. This chapter also discusses percent composition and different chemical formulas, including empirical formulas and molecular formulas. This wiki-page outlines this chapter in its entirety and includes information crucial to the understanding of these chapters. =====



**__﻿10.1 The Mole: A Measurement of Matter__**
__﻿__ __Katherine Perry - Page 287- 289 - Measuring Matter__ **Key Concept:** You often measure the amount of something by one of three different methods-by count, by mass, and by volume. Examples: --number of CDs you have --number of pins you knocked down in bowling --pounds of potatoes --kilograms of sugar --ounces of gold --grams of silver --liters of gasoline --gallons of milk --milliliter of medicine --teaspoon of baking powder Heather Bowditch - Picture

Some unites indicate a specific number of items. Using DA, you could calculate the mass of a bushel of apples. Examples: --pair = 2 --dozen = 12 --count of apples = 3 for $2.40 --pounds of apples = $1.29/pound --kilograms of apples = $2.79/kg --bushel of apples = $12.00/bushel

__Adam Shanahan - Page 290 - What Is a Mole?__ Courtney Gareau - Picture

A **mole** is is a unit used to measure the amount of substance of a certain material. A mole of a substance is 6.02 X 10^23 representative particles of a certain substance. The number of representative particles in a mole (6.02 X 10^23) is called **Avogadro's numbe**r. This is named after the Italian scientist Amedeo Avogadro di Quaregna (1776-1856) who clarified the difference between molecules and atoms. A **representative particle** refers to thespecies present within a substance. A mole of any substance contains Avogadro's number of representative particles, or 6.02 X 10^23 representative particles.

__ Christos Anastos - Pages 290-293 - What is a Mole?__ Mole (mol)- 6.02x10^23 representative particles of a substance, which is the SI unit for measuring the amount of a substance. The number of representative particles in a mole, 6.02x10^23, is called Avogadro’s Number.

- The number of representative particles in a mole. (6.02x10^23)
 * Avogadro’s Number **

- the species present in a substance Atoms Molecules Formula Units
 * Representative Particle **

A mole of any substance contains Avogadro’s number of representative particles. Because 1 mol=6.02x10^23 representative particles, that formula can be used as the basis for a conversion factor to help you convert numbers of representative particles to moles Moles= representative particles x 1 mol/6.02x10^23 representative particles The opposite is also true. It can be used as a conversion factor to help convert number of moles to the amount of representative particles Representative Particles=moles x 6.02x10^23 / 1 mole

media type="youtube" key="xiVweBpjXJo" height="390" width="480" Heather Bowditch - Youtube Video

__Nick Brault p.293-294: The Mass of a Mole of an Element__


 * Atomic mass is expressed in AMUs (atomic mass units), which is a relative weight system based on carbon
 * For example, the ratio of hydrogen’s atomic weight to carbon’s weight is 1:12. This means that any amount of carbon weighs twelve times as much as the same amount of hydrogen.
 * These atomic weights are not whole numbers because atomic mass is the average of the weights of isotopes of the element.
 * The mass of an element in grams is its molar mass.
 * For example 12.0g of carbon has the same amount of particles as 16.0g of carbon.

media type="youtube" key="hy0MX2BHNGg" height="390" width="480" Heather Bowditch - Youtube Video

__Andrew Scotti - Pages 295-296- The Mass of a Mole of a Compound__


 * Key Concept -** To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound. Then add the masses of the elements in the compound.
 * To find the mass of a mole of a compound you need to first know the formula of the compound.

Example: Sulfur trioxide (SO3)


 * 1. Break formula into each individual element ( 1 sulfur atom and 3 oxygen atoms)
 * 2. Look at periodic table and find the atomic mass of each element (sulfur = 32.1 amu and oxygen = 16.0 amu and there are 3 of them so 48.0 amu)
 * 3. Add all the masses together of each element to get the molecular mass (32.1 amu + 48.0 amu = 80.1 amu)
 * 4. Molar mass of compound is equal to the molecular mass of each atom of a compound (molar mass of SO3 = 80.1 g
 * 5. The molar mass is the mass of 6.02 x 1023 molecules of that compound

Overview: To find the mass of a mole of a compound you add the atomic mass of each part of a compound together and change amu to g and you get molar mass.
 * These calculations from finding molar mass works for any compound, whether it be molecular or ionic.

**__ ﻿10.2 Mole-Mass and Mole-Volume Relationships __**
__Mike McShane - Page 297-299 - The Mole-Mass Relationship__

We know that the molar mass of a substance is the mass (in grams) of one mole of the substance. This applies to all substances. There can be some confusion though. In terms of oxygen for example, it depends on what you assume to be the representative particle of oxygen. One person might think that it is O2 because oxygen cannot be found by itself and another person might think that it is just O. In the first case, the molar mass would be 32.0g, while in the second it would be 16.0g. The only way to avoid these kinds of mistakes is to, in the questions, use either O or O2.

You can use the molar mass of an element or compound to convert between the mass of a substance and the moles of a substance. You can do this using the equation: Mass (in grams) = the number of moles X mass (in grams)/1 mole. For example, if you need 3.00 mol of sodium chloride, it would be easy to mass the amount and in order to do this you must convert the 3.00 mol of NaCl into a mass in grams. First you must find out the molar mass of the substance which in this case it 58.5 g/mol. Then you plug the numbers into the equation: Mass of NaCl = 3.00 mol X 58.5g/1mol By using the equation, you can figure out that 3.00 moles of NaCl equals 176g of NaCl. This helps because now you can measure out 3.00 moles by measuring out 176g.

Sodium chloride -Mike McShane

__Tom DeMarco - Page 297-299 - The Mole-Mass Relationship __ To convert between the mass of a substance and the moles of a substance, use the molar mass of an element or compound. Sample Problems 1) Converting Moles to Mass: Number of moles= 9.45 mol Al2O3 <span style="font-family: Arial,Helvetica,sans-serif;">Mass= g Al2O3 <span style="font-family: Arial,Helvetica,sans-serif;">Mass= 9.45 mol Al2O3 x 102.0 g Al2O3/1 mol Al2O3 <span style="font-family: Arial,Helvetica,sans-serif;">= 964g Al2O3 <span style="font-family: Arial,Helvetica,sans-serif;">- Explanation: This answer makes sense because the number of moles of Al2O3 is about 10 and each has a mass of about 100 g; therefore, the answer must be near 1000 g. The answer has also been rounded to the correct number of significant figures.

Courtney Gareau - Diagram

<span style="font-family: Arial,Helvetica,sans-serif;">2) Converting Mass to Moles: <span style="font-family: Arial,Helvetica,sans-serif;">Mass= 92.2g Fe2O3 <span style="font-family: Arial,Helvetica,sans-serif;">Number of Moles=_ mol Fe2O3 <span style="font-family: Arial,Helvetica,sans-serif;">Moles= 92.2g Fe2O3 x 1 mol Fe2O3/ 159.6g Fe2O3 <span style="font-family: Arial,Helvetica,sans-serif;">= 0.578 mol Fe2O3 <span style="font-family: Arial,Helvetica,sans-serif;">- Explanation: This answer makes sense because the given mass is slightly larger than a half mole of Fe2O3, so the answer is a little bit greater than one half mol.

__Eileen Corkery- Pages 300-302- The Mole-Volume Relationship__

-The mole is a measuring device that is relative to what is being measured. For example, 1 mole of water is very different than 1 mole of glucose. - The volumes of moles of gasses that are measured under the same conditions are very predictable. -Part of the “mole theory” is based upon //Avogadro’s hypothesis// (the number of representative particles contained in one mole of a substance- also known as 6.02x 10^23) -Avogadro said that equal volumes of gasses at the same temperature and pressure contain equal numbers of particles, even though the particles are not all the same size. Courtney Gareau - Picture - As already stated, Avogadro’s hypothesis only works “at the same temperature and pressure”. Another way of saying this is //STP (Standard temperature and pressure)//; STP is a temperature of 0 degrees C and a pressure of 101.3 kPa (or 1 atmosphere- atm) - It is important for a gas to be tested at STP because of various changes that may occur. Some examples of this are when a balloon starts to deflate because of low temperatures, or a water bottle condenses in high pressures. -At STP, 1 mol (or 6.02x 10^23 representative particles) of any gas occupies a volume of 22.4 L. This 22.4 L is known as the //molar volume// of a gas.

media type="youtube" key="fexEvn0ZOpo" height="390" width="480" align="center"

Eileen Corkery - Youtube Video - This video explains Avogadro's hypothesis in more depth.

__Marion Burdick- Pages 300-302 - Equations__

Marion Burdick - Pictures

__Steven Denison- Page 303 - The Mole Road Map__ To convert from one unit to another the mole must be used as an intermediate step. Mass to Mole: <span style="font-family: Calibri,sans-serif; font-size: 15px; line-height: 22px;">1.00 mol/molar mass <span style="font-family: Calibri,sans-serif; font-size: 15px; line-height: 22px;">Mole to Mass: molar mass/1.00 mol

<span style="font-family: Calibri,sans-serif; font-size: 15px; line-height: 22px;">Volume of Gas to Mole: 1.00 mol/22.4 L <span style="font-family: Calibri,sans-serif; font-size: 15px; line-height: 22px;">Mole to Volume of Gas: 22.4 L/1.00 mol

<span style="font-family: Calibri,sans-serif; font-size: 15px; line-height: 22px;">Particles to Mole: 1.00 mol/6.02 x 10^23 particles <span style="font-family: Calibri,sans-serif; font-size: 15px; line-height: 22px;">Mole to Particles: 6.02 x 10^23 particles/1.00 mol

Steven Denison - Picture


 * __﻿10.3 Percent Composition and Chemical Formulas__**

__Nina DeMeo - Pages 305-306 - The Percent Composition of a Compound__ The Percent Composition of a Compound __Key Concept__: The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplies by 100% - knowing the relative amounts of the components of a mixture or compound is often useful - For example: knowing the appropriate amount of a certain nutrient needed in fertilizer - the relative amounts of the elements in a compound are expressed as the **percent composition** (a.k.a. the percent by mass of each element in the compound) - consists of a percent for each element in a compound - the percents must total 100%

Percent Composition from Mass Data - use analytical procedures to determine the masses of each element in a compound and calculate the percent composition **Example:** When a 13.60-g sample of a compound, containing only Mg and O, is decomposed, 5.40-g of O is obtained. What is the percent composition of this compound? Known: - mass of compound = 13.6 g - mass of O = 5.4 g  - mass of Mg = 8.2 g Unknown: - percent of Mg = ? % Mg - percent of O = ? % O -percent of Mg = 8.2g/13.6g × 100% = 60.3%

Percent of O = 5.4g/13.6g × 100% = 39.7%

Nina DeMeo - Picture

__ Evan Grandfield - Pages 307-308 - Percent Composition from the Chemical Formula __

media type="youtube" key="1CW6m3cwquE" height="325" width="388" Evan Grandfield - Youtube Video

I. Percent Composition from the Chemical Formula A. You can calculate the percent composition of a compound even if all you know is its chemical formula. 1. Subscripts in the chemical formula are used to calculate the mass of each element in a mole of that compound. a. Sum of these masses = molar mass 2. You can use the individual masses of the elements and the molar mass to calculate the % by mass of each element in 1 mole of the compound. 3. You do this by dividing the mass of each element by the molar mass. Then you multiply this result by 100%.

Way to Do It

% mass = [(mass of element in 1 mol compound) / (molar mass of compound)] X 100% *percent composition of a compound is always the same

Sample Problem

Calculate the percent composition of propane (C3­­H8).

We know that: 1. mass of C in 1 mol C3­­H8 = 36.0g 2. mass of H in 1 mol C3­­H8 = 8.0g 3. molar mass of C3­­H8 = 44.0g/mol

We need to know that: 1. percent C = ?%C 2. percent H= ?%H

You must now calculate by mass of each element by dividing the mass of that element in 1 mol of the compound by the molar mass of the compound. Then multiply by 100%.

The calculations should look like this:

%C = [(mass of C)/ (mass of C3­­H8)] X 100% = 36.0g/44.0g X 100% = 81.8% %H = [(mass of H)/(mass of C3­­H8)] X 8.0g/44.0g X 100% = 18%

II. Percent Composition as a Conversion Factor A. Percent Composition can be used to calculate the number of grams of any element in a specific mass of a compound.

Step 1: Multiply the mass of the compound by a conversion factor which is based off the percent composition of the element in the compound.

Follow this example, solving how much carbon and hydrogen are in 82.0g of C3­­H8­ ­(propane).

1. Propane is 81.8% carbon and 18% hydrogen. 2. In a 100g sample of propane (C3­­H8) there are 81.8g of carbon and 18g of hydrogen. 3. Use this ratio— 81.8gC/100g C3­­H8 to calculate mass of carbon in 82.0g of C3­­H8 4. Use this ratio— 18gH/100gC3­­H8 to calculate the mass of hydrogen

Do the Math!

82.0g C3­­H8 X [(81.8gC)/(100g C3­­H8)] = 67.1g C

82.0g C3­­H8 X [(18gH)/(100 g C3­­H8<span style="font-family: 'Times New Roman',serif; font-size: 16px;">)] = 15g H

Sum of the two masses = 82g which is the sample size to 2 sig figs

Evan Grandfield - Picture

__ Kendyl Barron - Pages 309-310 - Empirical Formula __

The empirical formula of chemical compounds shows a basic ratio of elements within that compound. Multiplying this ratio by any factor produces the formula of other compounds. The basic ratio of the elements in a compound can be calculated using the percent composition of each element in the compound. The basic ratio is also referred to as the empirical formula. Empirical formulas can be described on an atomic or molecular level. The molecular formula of a compound tells the actual number of each kind of atom in the compound. The empirical formula of a compound isn’t always the same as the molecular formula, as it is the lowest ratio of each element to one another. Empirical refers to the elemental analysis of relative amounts of elements in a compound. Elizabeth Sieber - Picture

Determining the Empirical Formula (ex): --> compound contains 25.9% nitrogen and 74.1% oxygen. --> percent (parts per 100) assume 100g of compound contains 25.9g of N and 74.1g of O. --> covert to moles: 25.9g N x 1 mol N/ 14.0g N = 1.85 mol N 74.1g O x 1 mol O/16.0g O = 4.63 mol O  --> Divide each molar quantity by the number of smaller moles to get whole number ratios. 1.85 mol N/1.85 = 1 mol N 4.63 mol O/1.85 = 2.5 mol O  -->multiply by 2 to get whole numbers: N2O5

__Maggie Bie - Pages 311-312 - Molecular Formulas__

<span style="font-family: Arial,Helvetica,sans-serif;">Key Concept – 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.

<span style="font-family: Arial,Helvetica,sans-serif;">Once you determine the empirical formula of your compound, you can determine its molecular formula but you need the molar mass.

<span style="font-family: Arial,Helvetica,sans-serif;">Molar mass found through a mass spectrometer: <span style="font-family: Arial,Helvetica,sans-serif;">Compound is broken into charged fragments (ions) that travel through a magnetic field <span style="font-family: Arial,Helvetica,sans-serif;">Magnetic field detects the particles from their straight-line path <span style="font-family: Arial,Helvetica,sans-serif;">Mass of compound is determined from the amount or deflection experienced by the particles

<span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;"> <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;">Maggie Bie - Picture

<span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;">From the empirical formula, you can find the empirical formula mass (efm) <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;">EFM = the molar mass represented by the empirical formula

<span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;">The number of empirical formula units in a molecule of the compound (also the multiplier to convert the empirical formula to the molecular formula) <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;">Divide the experimentally determined molar mass by the empirical formula mass <span style="font-family: 'Times New Roman',serif; font-size: 12pt; margin: 0in 0in 0pt;">To determine the molecular formula from the empirical formula, multiply the subscripts in the empirical formula.

media type="youtube" key="nslC7lOSc7Y" height="273" width="336" Maggie Bie - Youtube Video

__ Elizabeth Sieber - Sample Problems __ Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH4N.

Courtney Gareau - Picture of CH4N

Start with molar mass that is shown by the formula. (look at the periodic table to get these numbers) Now divided the molar mass given (60.0 g/mol) by the molar mass shown by the actual formula (30.6) to get a WHOLE NUMBER!!
 * C - 1 * 12.01 = 12.01
 * H - 4 * 1.01 = 4.04 (there are 4 hydrogens in CH4N, so I multiplied it by 4!!)
 * N - 1 * 14.01 = 14.01
 * add them all up - 30.06

Finally, "distribute" the number we just found (2) to the formula to get the molecular formula.
 * 60.0 / 30.06 = 1.996 (about 2, because it needs to be a whole number)
 * CH4N ---> C2H8N2 (Give yourself a pat on the back! You have now found a molecular formula! )