Chapter+10+The+Mole

= **__Chapter 10: The Mole__** =
 * In this chapter, we will be learning about the mole. No we will not be learning about the small, blind, underground-dwelling mamal. It is an important unit of measurement in chemistry. It measures the amount of stuff something has. It can also be defined as 6.02x10^23. This measurement is so important, that there is an ENTIRE day dedicated to it. Mole Day is celebrated on October 23rd.**



= Section 1 = = **__Measuring Matter__** = -chemistry is a quantitative science and we live in a quantitative world-alot of things are based off of numbers -analyze the composition of matter -perform chemical reaction to quantities of products -3 different ways to measure the amount of something - find mass - find volume -count (continued)
 * Some units used for measuring indicate a specific number of items
 * Dozen=12, Pair=2, Avogadro's number=6.02x10^23
 * ie: Apples
 * By count (3 for $4.30)
 * By weight ($1.29/lb)
 * By volume ($12.00 a bushel)
 * All these ways can be equated to a dozen apples
 * 1 dozen=12 apples
 * 1 dozen=2.0 kg apples
 * 1 dozen= 0.20 bushel apples
 * Knowing how the count, mass, and volume of apples relate to a dozen apples allows you to convert among these units
 * Based on the unit relationships, you could calculate the mass of a bushel of apples or the mass of 90 apples using **conversion factors:**
 * __1 dozen__ __2.0 kg__ __1 dozen__
 * 12 1 dozen 0.02 bushel

__**What is a Mole?**__
- matter is compsed of atoms, moelcules, ions which are all extremely small - eggs are counted by the dozen - chemists count the number of particles by the mole (abbreviated mol) - the number which represents the particles of that substance is 6.02 x 10^23 - this is called Avogadro's number - the term representative particle refers to the species present in a substance; for most elements it is the atom - a mol of any substance contains Avogadro's number of representative particles, or 6.2 x 10^23 representative particles this video does a nice job explaining the mole! []

**__Converting Number of Particles to Moles__**
1. Analyze and list the knowns and unknowns. 2. Calculate and solve for the unknown 3. Evaluate your answer and see if the result makes sense.
 * The relationship of 1 mole = 6.02 x 1023 is the basic conversion factor that converts numbers of representative particles to mole.
 * Sample Problem
 * How many moles of magnesium is 1.25 x 1023 atoms of magnesium?
 * **Knowns:**
 * number of atoms = 1.25 x 1023
 * 1 mole Mg = 6.02 x 1023 atoms Mg
 * Desired conversion is atoms- à moles
 * **Unknowns**
 * Moles = ? mole Mg
 * Conversion factor is 1 mole Mg/ 6.02 x 1023
 * Moles = 1.25 x 1023 atoms Mg X 1 mole Mg/ 6.02 x 1023
 * Moles = 2.08 x 10-1 mole Mg = .208 mole Mg

**__Converting Moles to Number of Particles__**

 * To determine how many atoms are in a mole of a compound, you need to know how many atoms are in a representative particle of the compound.
 * This number is determined by the chemical formula
 * For example in carbon dioxide the chemical formula is CO2, this subscript 2 shows that there is one carbon atom but two oxygen atoms.
 * A mole of carbon dioxide contains Avogadro’s number of CO2molecules. But each molecule contains three atoms.
 * Therefore 1 mole of CO2 = 3 x Avogadro’s number

**__Coverting Moles to Number of Particles Continued__**

 * to find the number of atoms, you must first determine the number of representative particles
 * Representative particles = moles X 6.02 X 10^23 representative particles/ 1 mole
 * Example:** How many atoms are in 2.12 mol of propane (C3H8)?
 * The first conversion factor is 6.02 X 10^23 molecules C3H8/ 1 mol C3H8
 * The second conversion factor is 11 atoms ( 3 carbon and 8 hydrogen atoms) / 1 molecule C3H8
 * Then multiply 2.12 mol C3H8 X (6.02 X 10^23 molecules C3H8/ 1 mol C3H8) X (11 atoms / 1 molecule C3H8)
 * The units cancel and the answer is 1.4 X 10^25 atoms

__**Finding the Molar Mass of a Compound**__

 * Example:** What is the molar mass of hydrogen peroxide (H2O2)?
 * 1) Analyze, solve for the unknown then check to see if the result makes sense?

grams of H= 2 mol H X (1.0 gH) / (1 mol H)= 2.0 grams of H grams of O= 2 mol O X (16 gO) / (1 mol O)= 32 grams of O

molar mass of H2O2= g of H+ g of O = **34 grams**
 * More examples:**



__**The Mass of a Mole of an Element**__
[] This helps to give a better idea of how the conversion works so easily and then goes on to show how to do a conversion that begins with a greater amount.
 * Atomic mass is expressed in atomic mass units (amu)
 * Chemists have converted the relative scale of masses of the elements in amu to a relative scale of masses in grams because they are easier to work with
 * **The atomic mass of an element expressed in grams is the mass of a mole of the element.**
 * Molar mass: the mass of a mole of an element
 * The molar masses of any two elements must contain the same number of atoms
 * The molar mass of any element contains 1 mol or 6.02 x 10^23 atoms of that element
 * Mole also can be described amount of substance that contains as many representative particles as the number of atoms in 12.0g of carbon-12

**__The Mass of a Mole of a Compound__**

 * To find the mass of a mole of a compound you first need to know the formula of the compound.
 * For example SO3 has one atom of Sulfur and three atoms of Oxygen.
 * Then to calculate the mass, you add the atomic masses of the atoms making up the molecule.
 * For example Sulfur has an amu (atomic mass unit) of 32.1. While there are three Oxygen atoms though so Oxygen’s amu, must be multiplied by 3: 16 x 3 = 48. So added together the molecular mass of So2 is 80.1 amu.
 * Now substitute the unit grams for amu to find the molar mass of SO3. The molar mass is the mass in grams of 1 mole of the compound.
 * For example 1 mole of SO3 has a mass of 80.1 grams.
 * This is the mass of 6.02 x 1023 molecules in SO3.
 * In conclusion to find molar mass, find the number of grams of each element in one mole of the compound. Then add the masses of the elements in the compound.
 * This method can be applied to any compound, molecular, or iconic.

= Section 2 = When guess how many Jelly Beans are in a jar,and you win, is it just a lucky guess? No! You estimate the length and diameter of the Jelly Beans to find it's estimated volume. Then you estimated the dimensions of the container. This is how you came up with such an educated guess. It is similar to how chemists use the relationships between moles and quantities like mass, volume, and number of particles to solve chemistry problems!
 * Connecting to Your World**

__The Mole-Mass Relationship__
1 mol
 * applies to all substances (elements, molecular compounds, ionic compounds)
 * molar mass depends on what you assume to be the representative particle
 * avoid confusion by using the formula of the substance
 * **use the** molar mass **of an element/compound to convert between the mass of a substance and the moles of a substance**
 * molar mass=1mol
 * mass (grams)= number of moles= __mass (grams)__

__Finding Mass from Moles and Moles from Mass__
Here is an example of how to find the mass of a substance in a certain number of it's moles


 * Find the mass of 9.45 mol Al2O3
 * To begin this problem, we first need to understand that our known is 9.45mol, which we then need to convert into mass.
 * We can do this by finding the mass of one mol of each element in this compound, and multiplying this mass by the number of moles of this element present in the compound.
 * Al= 2 moles x 26.98g=53.96g
 * O= 3 moles x 16.00g=48g
 * Next, add these together to get the mass of one mol of Al2O3
 * 53.96g+48g=101.96 g in 1 mole
 * After this we can begin to solve for the mass of 9.45 moles of this compound
 * mass = 9.45 mol Al2O3/1 x 101.96g/ 1 mole = 9.45x101.96=**__963.52 grams in 9.45 moles__**.

To do problems similar to this one, simply follow the formula: moles= mass (g) x 1 mol/ mass (g)

ex. The molar mass of Na2SO4 is 142.1 g/mol, how many moles in 10.0g?

moles of Na2SO4=10.0g/1 x 1 mol/142.1 g= 10.0/142.1=.__**07 moles**__

__Mole-Volume Relationship,__
--- A collection of relatively large particles in a gas does not require more space than the same number of relatively small particles
 * The volume of moles of liquids and solids are varying and inconsistent
 * The volume of moles of gases are much more predictable
 * **Avogadro’s Hypothesis** – Equal volumes of gases at the same temperature and pressure contain equal numbers of particles
 * The volume of a gas changes with a change in temperature, and with a change in pressure
 * Because of these variations, the volume of a gas is measured at a standard temperature and pressure
 * **Standard Temperature and Pressure (STP)** – A temperature of 0oC and a pressure of 101.3 kPa, or 1 atmosphere (atm)
 * At STP, 1 mol of 6.02 x 1023 representative particles, of any gas occupies a volume of 22.4 L.
 * 22.4 L is called the **molar volume** of gas

__The Mole Road Map__
Now you know the relationships between a mole particles, mass, and volume of gases at STP. The mole is at the center of your chemical calculations. To convert from one unit to another you must use the mole as an intermediate step. Here is a diagram to assist you.

Here's a video that could also help http://www.youtube.com/watch?feature=player_detailpage&v=mBVL0PHPrhg = Section 3 = = __The Percent Composition of a Compound__ =

//Percent Composition from Mass Data//
To do this... Take, for example, MgO. You know you have Magnesium and Oxygen. The molar mass of this compound is 13.60 g where 5.40 g is Oxygen and 8.20 g is Magnesium. To calculate percent composition, divide the mass of the element by the mass of the compound and then multiply by 100. Divide 5.40 g into 13.60 to get .397 and then multiply this by 100 to get 39.7% which is your percent composition. Do the same for Magnesium to get 60.3% For more help of percent composition, use this webiste: []
 * =====If you want to verify the composition of your new compound and determine its molecular formula, you can use analytical procedures to determine the relative masses of each element in the compound and calculate its percent composition.=====

**__Percent Composition from the Chemical Formula__**
 * Percent composition of a compound can be calculated if only the chemical formula is known
 * The subscripts in the formula of the compound are used to calculate the mass of each element in a mole of that compound
 * The sum of the masses in the compound is the molar mass
 * Using the individual masses of the elements and the molar mass, the percent of each element can be calculated by mass in one mole of the compound
 * Divide the mass of each element by the molar mass and multiply the result by 100%
 * % Mass = (mass of element in 1 mol compound)/ (molar mass of compound) x 100%
 * The percent composition of a compound is always the same
 * Example – Propane (C3H8), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane.
 * Mass of C in 1 mol C3H8 = 36.0g
 * Mass of H in 1 mol C3H8 = 8.0g
 * Molar mass of C3H8 = 44.0 g/mol
 * Need to find %C and %H
 * %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%
 * Check by doing 81.8% + 18% = 99.8% (close to 100%)



**__Percent Composition as a Conversion Factor__**

 * Percent composition can be used to calculate the number of grams of any element in a specific mass of a compound
 * Multiply mass of the compound by a conversion factor based on the percent composition of the element in the compound
 * To find how much carbon and hydrogen are contained in 82.0g of propane, use the percent composition found in the example above
 * Carbon is 81.8% and Hydrogen is 18%, so in 100g sample of propane, would have 81.8g of carbon and 18g of hydrogen
 * Use ratio of 81.8g C/ 100g C3H8 to calculate the mass of carbon contained in 82.0g of propane (C3H8)
 * 82.0g C3H8 x (81.8g C)/ (100g C3H8) = 67.1g C
 * Using ratio of 18g H/ 100g C3H8, can calculate the mass of hydrogen
 * 82.0g C3H8 x (18g H)/ (100g C3H8) = 15g H
 * 67.1g C + 15g H = 82.1g C3H8 (close to 82)

by Kyle Gallagher

__Empirical Formulas__ Start with the number of grams of each element, given in the problem. Convert the mass of each element to moles using the molar mass from the periodic table. Divide each mole value by the smallest number of moles calculated. Round to the nearest whole number. This is the mole ratio of the elements and is represented by subscripts in the empirical formula.
 * The formulas for compounds are a basic ratio of elements
 * One can multiply by a common factor and a formula for a new compound will be formed
 * **Empirical Formula:**A basic ratio, which gives the lowest whole number ratio of the atoms of the elements in a compound
 * Shows the kinds and counts of atoms or moles
 * May or may not be the same as the molecular formula
 * The Empirical formula of a compound shows the smallest whole-number ratio of the atoms in the compound
 * Whereas the molecular formula shows the actual number of each kind of atom present in a molecule of the compound
 * Example Problem #1** **A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H. What is the empirical formula of the compound?**

__**Molecular Formulas**__
= = Example: Empirical Formula for hydrogen peroxide is HO. 17.0 g/mol
 * some compounds have the same empirical formula
 * Methanal, ethanoic acid and glucose = CH2O
 * 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
 * to find the molar formula you need both the empirical formula and the compound's molar mass
 * some scientists use a mass spectrometer to determine molar mass
 * compound is broken down into ions that travel through a magnetic field which deflects the particles from their straight-lined paths.
 * from empirical formulas, it is possible to calculate the empirical formula mass (emf)
 * this is the molar mass represented by the empirical formula
 * divide the the experimentally determined molar mass by the empirical formula mass which equals the number of empirical formula units in a molecule
 * it's empirical formula mass is 17.0 g/mol
 * The molar mass of H2O2 is 34.0g/mol
 * __34.0 g/mol__ = 2
 * the get the molecular formula of hydrogen peroxide from its empirical formula, multiply the subscipts in the empirical formula by 2
 * (HO) x 2 = H2O2