Q: How long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second? A:

: The Mole- 6.02 x 10 23 ODE TO A MOLE I find that my heart beat goes out of control Just thinking how useful to man is the mole! So perfectly compact. What could be neater? Only occupying twenty-two and four-tenths of a litre. What kind of equations could one hope to equate Without calculating a formula weight? And could a solution be kept for posterity Without ever knowing its molarity? Though all of these findings may set you to slumber I myself am aroused by Avogadro's number. Life without the mole? Don't be absurd! Count your blessings up to six point zero two times ten to the twenty-third. Mollionaire Q: How long would it take to spend a mole of $1 coins if they were being spent at a rate of 1 billion per second? A: The Mole A counting unit. Similar to a dozen (12), gross (144), pair (2), and ream (500), except instead of 12, 144, 5 or 500, it is 602 billion trillion, 602,000,000,000,000,000,000,000; 6.02 X 10 23 (in scientific notation) This number is named in honor of Amedeo (1776 1856), who studied quantities of gases and discovered that no matter what the gas was, there were the number of molecules present. A Mole of Particles contains 6.02 x 10 23 particles 1 mole C = 6.02 x 10 23 C atoms 1 mole H 2 O = 6.02 x 10 23 H 2 O molecules 1 mole NaCl = 6.02 x 10 23 NaCl formula units (technically, ionics are compounds not molecules so they are called formula units) = 6.02 x 10 23 Na + ions and = 6.02 x 10 23 Cl ions Atomic Mass It is useful to associate atomic mass with a mass in grams. It has been found that: 1 g H-1, 12 g C-12, or 23 g Na-23 have atoms 6.02 x 10 23 is a mole or mol is used in equations, mole is used in writing; one gram = 1 g, one mole = 1 mol. Since the mole is so large, we use it to count very tiny things like atoms. Because the mole is so large, (and we now know that we cannot count out a mole of anything), how do we know when we have a mole of anything? H1

: We the mass and that to the number of atoms present. Molar Mass The molar mass is the mass of mole of a pure substance. The pure substance can be an element or a compound. The atomic mass is the mass of atom of that element measured in amu s. The molar mass is also equal to 1 mole of atoms measured in grams. Comparing sugar (C 12 H 22 O 11 ) & H 2 O Do they have the same 1 g of each 1 mole of each volume? mass? # of moles? # of molecules? # of atoms Atomic Mass vs. Molar Mass Mass of 1 atom of Pb = amu Mass of 1 mole of Pb atoms = g Mass of 1 atom of N = amu Mass of 1 mole of N atoms = g Molecular Mass/Molecular Weight (Molecular Compounds): If you have a single molecule, mass is measured in instead of grams. But, the molecular mass/weight is the numerical value as 1 mole of molecules. Only the units are different. Formula Mass/Formula Weight (Ionic Compounds): Same goes for ionic compounds. But again, the numerical value is the same. Only the units are different. Molar Mass for Compounds How to find the molar mass: 1. Write a CORRECT formula for the compound 2. Look up the atomic mass of each element in the compound 3. Multiply the atomic mass by the subscripts, if any. 4. Add all masses of elements together and use the unit, g/mol Example: Find the molar mass of copper (II) bromate. H2

: Practice: Calculate molar masses (to 2 decimal places) of: (NH 4 ) 2 CO 3 O 2 Conversions: The factor- label method (TEXTBOOK, P. 652) To use this we need: 1. desired quantity, 2. given quantity, 3. conversion factor The steps to follow Now we are ready to solve problems using the factor label method. The steps involved are: 1. Write down the desired quantity/units 2. Equate the desired quantity to given quantity 3. Determine what conversion factors you can use (both universal and question specific) 4. Multiply given quantity by the appropriate conversion factors to eliminate units you don t want and leave units you do want 5. Complete the math Factor-label Method example: Q - How many moles are present in 35.4 grams of Cu? H3

Percentage Composition Q: If you have a box containing 100 golf balls and 100 Ping-Pong balls, which type of ball contributes the most to the mass of the box? A: The same principle applies to finding the % composition of a compound. Different elements have different masses and this must be taken into consideration. Percent Composition (by mass) Percent Composition (by mass): Identifies the elements present in a compound as a mass percent of the total compound mass. The mass percent is obtained by the mass of each element by the Mass of a compound and to percentage. % composition of an element = mass of element mass of compound x 100% 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. Example: What is the percent composition of each element in NH 4 OH by mass? Practice: 1. Find the percentage composition by mass of aluminum thiocyanate. (Al- 13.41%, S- 47.80%, C-17.91%, N- 20.88%) 2. A student prepares a compound of tungsten chloride from 3.946 g of tungsten and 3.806 g of chlorine. Assuming the reaction goes to completion, calculate the percent composition. (W- 50.50%, Cl-49.50%) 3. How many grams of sodium will combine with 567.0 g of sulfur to form Na 2 S? (Total Mass= 1379.56 g, Mass of Na= 812.56g) H4

Empirical Formula The ratio of elements in a compound. It uses the possible number ratio of atoms present in a formula unit of a compound. If the percent composition is known, an empirical formula can be calculated. Compound Formula Empirical Formula Hydrogen peroxide Benzene Ethylene Propane Simplest and molecular formulae Chemical formulas are either simplest (a.k.a. empirical ) or molecular. Ionic compounds are expressed as formulas. Covalent compounds can either be molecular formulas (i.e. ) or simplest (e.g. ) Q - Write simplest formulas for propene (C 3 H 6 ), C 2 H 2, glucose (C 6 H 12 O 6 ), octane (C 8 H 14 ) A: Q - Identify these as simplest formula, molecular formula, or both H 2 O, C 4 H 10, CH, NaCl. A: How to find the empirical formulae of a compound: Calculations to find the simplest formula incorporate this rhyme: % to mass Mass to mole Divide by small Multiply till whole A chart form may help to organize work. Example 1: A compound contained 29.08 % Na, 40.58 % S, and 30.34 % O. Find the empirical formula of this compound. species mass or % molar mass mass or % n = molar mass divide by smallest from column 4 reduce to smallest whole number ratio H5

Example 2: A 5.72 g sample of washing soda (Na 2 CO 3. x H 2 O) is heated to give 2.12 g of anhydrous Na 2 CO 3. What is the simplest formula of the hydrated salt? species mass or % molar mass mass or % n = molar mass divide by smallest from column 4 reduce to smallest whole number ratio Example 3: A compound contained 40.0g C, 6.71g H, and 53.3g O. Find its empirical formula and the empirical formula mass. species mass or % molar mass mass or % n = molar mass divide by smallest from column 4 reduce to smallest whole number ratio Empirical Formula Mass = Mole ratios and simplest formula: Given the following mole ratios for the hypothetical compound AxBy, what would x and y be if the mole ratio of A and B were: Formula: A = 1 mol, B = 2.98 mol A = 1.337 mol, B = 1 mol A = 2.34 mol, B = 1 mol A = 1 mol, B = 1.48 mol If any result from Step 3 is a mixed number, you must multiply ALL values by some number to make it a whole number. Ex: 1.33 x 3; 2.25 x 4; 2.50 x 2, etc. Formulas for Compounds Empirical Formula possible set of subscript numbers Smallest number ratio All ionic compounds are given as formulas Molecular Formula The formulas of molecules It shows of the atoms present in a molecule It may be the as the E.F. or a whole number of its E.F. H6

molar mass ( g / mol) n = empirical formula mass ( g / mol) n represents a number multiplier from 1 to as large as necessary How to find the molecular formulae of a compound: 1. Calculate the empirical formula and the mass of the empirical formula 2. Divide the given molecular mass by the calculated empirical mass 3. Answer is a whole number multiplier 4. Multiply the empirical formula by the multiplier Example: Lactic acid has a molar mass of 90.08 g and has this percent composition: 40.0% C, 6.71% H, 53.3% O. What is the empirical and molecular formula of lactic acid? Assume a 100.0 g sample size. Step 1. Use Chart to find the Empirical Formula: species mass or % molar mass mass or % n = molar mass divide by smallest from column 4 reduce to smallest whole number ratio Step 2. Obtain the mass of the Empirical Formula: Step 3. Obtain the value of n (whole number multiplier): Step 4. Multiply the empirical formula by the multiplier Questions: 1. What information must be known to determine a) the empirical formula of a substance? b) the molecular formula of a substance? 2. Determine the molecular formula for each compound below from the information listed. substance simplest formula molar mass(g/mol) naphthalene C 5 H 4 128 H7

Part 1: Stoichiometry (Mass to Moles) 2 Mg (s) + O 2(g) 2 MgO (s) Note that the Law of Conservation of Mass is always obeyed. Consider: 4NH 3 + 5O 2 6H 2 O + 4NO In words, this tells us that for every moles of NH 3, moles of O 2 are required Is 4 g NH 3 / 5 g O 2 a conversion factor? Stoichiometry questions (1) -Factor-Label Method Consider: 4NH 3 + 5O 2 6H 2 O + 4NO 1. How many moles of H 2 O are produced if 0.176 mol of O 2 are used? 2. How many moles of NO are produced in the reaction if 17.00 mol of H 2 O are also produced? 3. How many grams of H 2 O are produced if 1.9 mol of NH 3 are combined with excess oxygen? 4. How many grams of O 2 are required to produce 0.3 mol of H 2 O? 5. How many grams of NO is produced if 12 g of O 2 is combined with excess ammonia? H8

Moving along the stoichiometry path We always use the same type of information to make the jumps between steps: grams (x) moles (x) moles (y) grams (y) Example: Consider: 4NH 3 + 5O 2 6H 2 O + 4NO a) How many moles of H 2 O can be made using 0.50 mol NH 3? b) What mass of NH 3 is needed to make 1.50 mol NO? c) How many grams of NO can be made from 120.0 g of NH 3? The steps to solving stoichiometric problems are as follows- CHART METHOD 1. Write the chemical equation 2. Balance the chemical equation 3. Write the molar ratio for the equation 4. Write all given masses of substances in the equation 5. Write the molar masses for all substances 6. Find the number of moles of each substance 7. Find the new molar ratio 8. Solve for the unknown Example: NH 3? Given 4NH 3 + 5O 2 6H 2 O + 4NO How many grams of NO can be made from 120.0 g of Molar Ratio(MR) 4 NH 3 5 O 2 6 H 2 O 4 NO Mass (m) Molar Mass (MM) Moles (n) H9

Stoichiometry- Limiting Reagents A balanced chemical equation indicates the number of of each REACTANT that will react and the number of of each PRODUCT that will be produced in the reaction Even if one reactant is present in excess only the amount to react, dictated by the molar ratio, will actually react. The amount reacting will be determined by the reactant that is in the amount ( ). Problems of this type may be recognized by the fact that information will be given about at two reactants. Q - How many moles of NO are produced if mol NH 3 are burned in mol O 2? 4 mol NH 3, 5 mol O 2 4 mol NH 3, 20 mol O 2 8 mol NH 3, 20 mol O 2 In limiting reagent questions we use the limiting reagent as the given quantity and the reagent that is in excess. Example: How many grams of NO are produced if 4 moles NH 3 are burned in 20 mol O 2? Solving Limiting Reagents: grams to moles Q - How many g NO are produced if 20 g NH 3 is burned in 30 g O 2? 1. First we need to calculate the number of moles of each reactant Molar Ratio What we Have Limiting Factor Test NH 3 O 2 The value indicates the Limiting Reactant Determine the reactant in excess and the excess amount if 20 g NH 3 is burned in 30 g O 2? Reactant NH 3 O 2 Molar Ratio What we have Limiting Factor Test What we need Excess H10

Stoichiometry (given = limiting) 1) Express all chemical quantities as moles 2) Determine the limiting reagent via a chart 3) Perform the stoichiometry using the limiting reagent as the given quantity Example: How many g NO are produced if 20 g NH 3 is burned in 30 g O 2? Given: 4NH 3 + 5O 2 6H 2 O + 4NO 4NH 3 5O 2 6H 2 O 4NO Molar Ratio (MR) Mass (m) Molar Mass (MM) Moles (n) PERCENTAGE YIELD (P. 336-338) Yield: Theoretical yield: Actual yield: Percent Yield = Q: Give 4 possible reasons why the actual yield in a chemical reaction often falls short of the theoretical yield. H11

Sample problem: What is the % yield of H 2 O if 138 g H 2 O is produced from 16 g H 2 and excess O 2? [143 g; 96.7%] Step 1: Write the balanced chemical equation: Step 2: Determine actual and theoretical yield. Actual is given, theoretical is calculated: Step 3: Calculate % yield: Practice problem: What is the % yield of NH 3 if 40.5 g NH 3 is produced from 20.0 mol H 2 and excess N 2? [227 g, 17.8%] Practice problem: When 5.00 g of KClO 3 is heated it decomposes according to the equation: 2KClO 3 2KCl + 3O 2 a) Calculate the theoretical yield of oxygen. [1.96 g] b) Give the % yield if 1.78 g of O 2 is produced. [90.9%] c) How much O 2 would be produced if the percentage yield was 78.5%? [1.537 g O 2 ] H12