Chemical Quantities & Stoichiometry Chapters 10 & 12 Chemical Quantities & Stoichiometry

How would you describe how much of a substance you have? By mass By volume By counting the # of atoms/molecules

The Mole Mole (mol): SI unit for measuring the amount of a substance. We can use a word like a “dozen” to specify a certain quantity. Mole (mol): SI unit for measuring the amount of a substance. 1 mol = 6.02 x 1023 representative particles Avogadro’s Number: 6.02 x 1023 Representative Particle: smallest unit that has all the characteristics of that substance.

What is the representative particle?  Element (ex. Cu): ___________________ Exception: The representative particle of the 7 diatomic elements is a molecule. (ex. H2) Covalent compound (ex. H2O): ____________ Ionic Compound (ex. NaCl): _____________ Atom Molecule Formula Unit (Molecule)

Conversions 4 moles Ca 6.02 x 1023 atoms Ca 1 mole Ca 4 moles Ca = atoms Ca. 4 moles Ca 6.02 x 1023 atoms Ca 1 mole Ca = 2.41 x 1024 atoms Ca

Conversions 5 x 1018 atoms Cu 1 mole Cu 6.02 x 1023 atoms Cu 5 x 1018 atoms Cu = moles Cu. 5 x 1018 atoms Cu 1 mole Cu 6.02 x 1023 atoms Cu = 8.3 x 10-6 moles Cu

Conversions 9.2 moles 6.02 x 1023 molecules F2 1 mole 9.2 moles F2 = molecules F2? 9.2 moles 6.02 x 1023 molecules F2 1 mole = 5.5 x 1024 molecules F2

Conversions = 1.1 x 1025 atoms F 9.2 moles F2 = atoms F? 9.2 moles F2 6.02 x 1023 molecules F2 2 atoms F 1 mole 1 molecule F2 = 1.1 x 1025 atoms F

Conversions = 1.22 x 1025 atoms 3.4 moles C2H4 = total atoms? 3.4 moles C2H4 6.02 x 1023 molecules C2H4 6 atoms 1 mole C2H4 1 molecules C2H4 = 1.22 x 1025 atoms

Molar Mass Molar Mass: The mass of one mole of an element or compound. Molar mass of a compound = the sum of the masses of the atoms in the formula Use the atomic masses in grams on the periodic table. Find the molar mass: Sr MgBr2 Ba3(PO4)2 = 87.62 grams/mol 24.3 + (2x 79.9) = 184.1 grams/mol Ba = 3 x 137.38 g P = 2 x 30.97 g O = + 8 x 16 g = 602.08 grams/mol

Conversions = 1.22 x 1025 atoms 3.4 moles C2H4 = total atoms? 3.4 moles C2H4 6.02 x 1023 molecules C2H4 6 atoms 1 mole C2H4 1 molecules C2H4 = 1.22 x 1025 atoms

Gram-Mole Conversions 1 mol = molar mass (in grams) 68 grams F2 = moles F2? 68 grams F2 1 mole F2 38 grams 68 / 38 = 1.8 moles F2

STP Standard Temperature and Pressure (STP): 0oC, 1 atm See Reference Tables

Molar Volume of a Gas Avogadro’s Hypothesis: equal volumes of gases at the same temperature and pressure contain equal numbers of particles. At STP, 1 mole of any gas occupies a volume of 22.4 L.

Mole-Volume Conversions 1 mol = 22.4 L at STP (gases only!!!) 5.4 moles He = L He at STP? 5.4 moles He 22.4 L He 1 mole He 5.4 x 22.4 = 120.96 L He

Mole-Volume Conversions 5.4 moles CH4 = L CH4 gas at STP? 5.4 moles CH4 22.4 L CH4 1 mole CH4 5.4 x 22.4 = 120.96 L CH4

Volume-Mole Conversion 560 L SO3 = mol SO3 560 L SO3 1 mole SO3 22.4 L SO3 560 / 22.4 = 25 mole SO3

Taking it to the next Level… Molar Mass 6.02 x 1023 particles 22.4 L at STP (gases only) 1 mole

Molar Mass-Density Conversions How would you do this???   grams  grams liters mole Example:  A gaseous compound composed of sulfur and oxygen has a density of 3.58 g/L at STP. What is the molar mass of this gas? (Density) (Molar Mass) 3.58 g 22.4 L L 1 mole 3.58 x 22.4 = 80.3 g/mole

Molar Mass-Density Conversion What is the density of krypton gas at STP? 83.8 grams Kr 1 mole mole 22.4 Liters 83.8 / 22.4 = 3.74 g/L Kr

Percent Composition by Mass Law of Definite Proportions: In samples of any chemical compound, the masses of the elements are always in the same proportion. => Allows us to write chemical formulas. Percent Composition by Mass Worksheet

Percent Composition x 100 Whole Percent Composition - % by mass of each element in a compound Percent = Part Whole x 100

Percent Composition Mass of 1 element Mass of compound x 100 % Al: 342 Example: Find the mass percent composition of Al2(SO4)3 x 100 54 342 x 100 = 15.8% % Al: Al: 2 x 27 = 54 S: 3 x 32 = 96 O: 12 x 16 = 192 Total Comp. = 342 96 342 % S: x 100 = 28.1% 192 342 %O: x 100 = 56.1%

Finding Chemical Formulas from Percent Composition # grams in 100 grams How many moles of each element? Divide by smallest #moles 18.8% Na 29.0% Cl 52.2% O Empirical Formula: ___________________

Empirical Formula Empirical Formula: lowest whole-number ratio. The formula for an ionic compound will always be the empirical formula. The formula for a covalent compound will not always be the empirical formula.

Molecular Formula Molecular Formula: either the same as the empirical formula (as for ionic compounds) or a simple whole-number multiple of the empirical formula.

To Calculate Empirical Formula Calculate the empirical formula of a compound containing 0.90g Ca and 1.60g Cl. Step 1: Convert GRAMS to MOLES. Ca: 0.90g 1 mole = 0.0224 mole Ca 40.1 g Cl: 1.60g 1 mole = 0.0451 mole Cl 35.5 g

Calculating Empirical Formula Step 2: DIVIDE the # of moles of each substance by the smallest number to get the simplest mole ratio. Ca: 0.0224 = 1 Cl: 0.0451 = 2.01 ~ 2 0.0224 0.0224 CaCl2

Calculating Empirical Formulas Step 3: If the numbers are whole numbers, use these as the subscripts for the formula. If the numbers are not whole numbers, multiply each by a factor that will make them whole numbers. Look for these fractions: 0.5  x 2 0.33  x 3 0.25  x 4

Molecular Formula Example Suppose the mass percents of a compound are 40% carbon, 6.70% hydrogen, and 53.3% oxygen. Determine the empirical formula for this compound. Since this compound is covalent, the actual formula may not be the simplest ratio of elements. If the molar mass of the compound is experimentally shown to be 90.0 g/mol, what is the molecular formula of this covalent compound?

Determining Molecular Formula Find the molecular formula of ethylene glycol (CH3O) if its molar mass is 62 g/mol. Step 1: CH3O = (12) + (3 x 1) + (16) = 31 Step 2: 62 / 31 = 2 Step 3: 2 (CH3O)  C2H6O2

Empirical/Molecular Example The percent composition of methyl butanoate is 58.8% C, 9.8% H, and 31.4 % O and its molar mass is 102 g/mol. What is its empirical formula? What is its molecular formula?

Answer to Emp/Mol Example 58.8% C 1 mole C = 4.9 / 1.9 = 2.5 x 2 = 5 12 g C 9.8% H 1 mole H = 9.8 / 1.9 = 5 x 2 = 10 1 g H 31.4%O 1 mole O = 1.9 /1.9 = 1 x 2 = 2 16 g O C5H10O2  (5x12) + (10x1) + (2x16) = 102 g/mol Empirical mass = molecular mass, so molecular formula is the same  C5H10O2

Warm Up: If a compound is 40% C, 7% H, and 53% O, what is its empirical formula? What is the molecular formula for this element if the molecular mass is 180 g/mol?

Stoichiometry Stoichiometry: The calculation of quantities of substances involved in chemical reactions. N2 (g) + 3H2 (g)  2NH3 (g)   The above equation could be read: 1 mol of N2 reacts with 3 moles of H2 to yield 2 moles of ammonia.

Mole-Mole Conversions 2A + B  3C + 7D Given the number of moles of reactant A (ex. 6 moles A), I can find: 1) The number of moles of reactant B needed to react completely with 6 moles of A (all 6 moles are used up). 2) The number of moles of product C formed. 3) The number of moles of product D formed.

Mole-Mole Examples N2 (g) + 3H2 (g)  2NH3 (g) There is a 1:3:2 mole ratio If you have 2 moles of N2, how many moles of NH3 will be produced? If you want 5 moles of product, how many moles of hydrogen gas do you need? How many moles of nitrogen are needed to react completely with 8 moles of hydrogen?

Mole to Mole Example 2 mol N2 2 1 mol NH3 mol N2 N2 + 3H2  2NH3 If you have 2 moles of N2, how many moles of NH3 will be produced? 2 mol N2 2 1 mol NH3 mol N2 2 x 2 / 1 = 4 moles NH3

Mole to Mole Example 5 mol NH3 3 2 mol H2 mol NH3 N2 + 3H2  2NH3 If you want 5 moles of product, how many moles of hydrogen gas do you need? 5 mol NH3 3 2 mol H2 mol NH3 5 x 3 / 2 = 7.5 moles H2

Mole to Mole Example 8 mol H2 1 mole N2 3 mole H2 N2 + 3H2  2NH3 How many moles of nitrogen are needed to react completely with 8 moles of hydrogen? 8 mol H2 1 mole N2 3 mole H2 8 x 1 / 3 = 2.67 mole N2

Reaction Conversions ****The only way to convert from one compound to something totally different in the reaction is to use the MOLE TO MOLE RATIO from the coefficients!!!****

Reaction Conversions 22.4 L at STP 1 mole Note – If you don’t have moles already, your first step is to convert to moles! Mole Review: Molar Mass 6.02 x 1023 particles 22.4 L at STP 1 mole

C. Molar Volume at STP (22.4 L/mol) (g/mol) particles/mol LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION

Volume-Volume Conversions How many liters of oxygen are required to burn 3.86 L of carbon monoxide? 2CO (g) + O2 (g)  2CO2 (g) How many liters of PH3 are formed when 0.42 L of hydrogen reacts with phosphorus? P4 (s) + 6 H2 (g)  4 PH3 (g)

Mass-Mass Conversions N2 (g) + 3H2 (g)  2NH3 (g) How many grams of hydrogen gas are required for 3.75 g of nitrogen gas to react completely? What mass of ammonia is formed when 3.75 g of nitrogen gas react with hydrogen gas?

More Mass-Mass Conversions N2 + 3H2  2NH3 How many grams of H2 are required to produce 5.0 grams of NH3? Grams NH3  moles NH3  moles H2  grams H2 5.0 g NH3 1 mole NH3 3 mole H2 2 g H2 17 g NH3 2 mole NH3 1 mole H2 5.0 x 3 x 2 / 17 / 2 = 0.88 g H2

More Practice* The combustion of propane, C3H8, a fuel used in backyard grills and camp stoves, produces carbon dioxide and water vapor. C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (g) What mass of carbon dioxide forms when 95.6 g of propane burns? Solid xenon hexafluoride is prepared by allowing xenon gas and fluorine gas to react. Xe (g) + 3F2 (g)  XeF6 (s) How many grams of fluorine are required to produce 10.0 g of XeF6? How many grams of xenon are required to produce 10.0 g of XeF6?

Mixed Practice - Examples How moles of CO2 are produced when 52.0 g C2H2 burns? 2C2H2 (g) + 5O2 (g)  4CO2 (g) + 2H2O (g) How many liters of hydrogen gas are formed from 50 grams of potassium? 2K (s) + 2H2O (l)  2KOH (aq) + H2 (g)

Mixed Practice – (Warm Up) How many molecules of oxygen are produced by the decomposition of 6.54 g of potassium chlorate (KClO3)? 2KClO3 (s)  2KCl (s) + 3O2 (g) How many grams of nitrogen dioxide must react with water to produce 5.00 x 1022 molecules of nitrogen monoxide? 3NO2 (g) + H2O (l)  2HNO3 (aq) + NO (g)

Limiting Reagent Limiting Reagent: The reactant that limits the amount of product that can be formed in a reaction. The reaction will stop when all of this reactant is used up. Determines the amount of product that is produced. Excess Reagent: You have more than you need of this reactant. The reaction will stop before all of this reactant is used up. You will have some of this reactant leftover.

Everyday Example You have : 1 loaf of bread (containing 14 slices of bread) 4 jars of peanut butter 2 jars of jelly A) How many peanut butter and jelly sandwiches can you make? B) What is the limiting reagent? C) What are the reactants in excess?

Limiting Reactants **The amount of product that can be formed in a reaction is always determined by the limiting reactant!!**

S’mores A s’mores MUST have: If you had: 2 graham crackers 2 pieces of chocolate 1 marshmallow If you had: 8 graham crackers 4 pieces of chocolate 6 marshmallows How many s’mores could you make?

Limiting Reagent Example Problems 1) If 2.70 mol C2H4 is reacted with 6.30 mol O2, what is the limiting reagent? C2H4 (g) + 3O2 (g)  2CO2 (g) + 2H2O (g)

(cont’d) 2) Identify the limiting reagent when 6.00 g HCl reacts with 5.00 g Mg. Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g) - How many grams of MgCl2 are produced in this reaction? __________________ Which reactant is in excess? _________________ How much of your excess reagent do you have leftover? _____________________

(cont’d)* 3) How many grams of water can be produced by the reaction of 2.40 mol C2H2 with 7.4 mol O2? C2H4 (g) + 3O2 (g)  2CO2 (g) + 2H2O (g) Limiting Reagent? ______________ Excess Reagent? _______________

Percent Yield Percent Yield = Actual Yield x 100% Theoretical Yield Theoretical Yield: the maximum amount of product that could be formed from the given amounts of reactants (ideal conditions). Calculated using stoichiometry. Actual Yield: the amount of product that actually forms in a lab. Actual yield is usually less than theoretical yield. Percent Yield = Actual Yield x 100% Theoretical Yield

Percent Yield Example Problem If 3.75 g of nitrogen completely react, what is the theoretical yield of NH3? N2 (g) + 3H2 (g)  2NH3 (g) If the actual yield is 3.86 g, what is the percent yield?

Practice* (WARM UP) Find the percent yield if 84.8 g of iron (III) oxide reacts with an excess of carbon monoxide to produce 55.0 g of iron. Fe2O3 (s) + 3CO (g)  2Fe (s) + 3CO2 (g)