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Chemical Quantities & Stoichiometry By mass By volume By counting the # of atoms/molecules ◦ 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. Atom Element (ex. Cu): ___________________ ◦ Exception: The representative particle of the 7 diatomic elements is a molecule. (ex. H2) Molecule Covalent compound (ex. H2O): ____________ Ionic Compound (ex. NaCl): Formula Unit _____________ (Molecule) 4 moles Ca = 4 moles Ca atoms Ca. 6.02 x 1023 atoms Ca 1 mole Ca = 2.41 x 1024 atoms Ca 5 x 1018 atoms Cu = Cu. 5 x 1018 atoms Cu moles 1 mole Cu 6.02 x 1023 atoms Cu = 8.3 x 10-6 moles Cu 9.2 moles F2 = 9.2 moles molecules F2? 6.02 x 1023 molecules F2 1 mole = 5.5 x 1024 molecules F2 9.2 moles F2 = 9.2 moles F2 6.02 x 1023 molecules F2 1 mole = 1.1 x 1025 atoms F atoms F? 2 atoms F 1 molecule F2 3.4 moles C2H4 = atoms? 3.4 moles C2H4 6.02 x 1023 molecules C2H4 1 mole C2H4 = 1.22 x 1025 atoms total 6 atoms 1 molecules C2H4 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: 1. Sr = 87.62 grams/mol 2. MgBr2 3. Ba3(PO4)2 24.3 + (2x 79.9) = Ba = P = O =+ = 184.1 grams/mol 3 x 137.38 g 2 x 30.97 g 8x 16 g 602.08 grams/mol 3.4 moles C2H4 = atoms? 3.4 moles C2H4 6.02 x 1023 molecules C2H4 1 mole C2H4 = 1.22 x 1025 atoms total 6 atoms 1 molecules C2H4 1 mol = molar mass (in grams) 68 grams F2 = 68 grams F2 moles F2? 1 mole F2 38 grams 68 / 38 = 1.8 moles F2 Standard Temperature and Pressure (STP): 0oC, 1 atm ◦ See Reference Tables 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. 1 mol = 22.4 L at STP (gases only!!!) 5.4 moles He = 5.4 moles He L He at STP? 22.4 L He 1 mole He 5.4 x 22.4 = 120.96 L He 5.4 moles CH4 = 5.4 moles CH4 L CH4 gas at STP? 22.4 L CH4 1 mole CH4 5.4 x 22.4 = 120.96 L CH4 560 L SO3 = 560 L SO3 mol SO3 1 mole SO3 22.4 L SO3 560 / 22.4 = 25 mole SO3 22.4 L at STP (gases only) 1 mole Molar Mass 6.02 x 1023 particles How would you do this??? grams (Molar Mass) (Density) 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? 3.58 g L 22.4 L 1 mole 3.58 x 22.4 = 80.3 g/mole What is the density of krypton gas at STP? 83.8 grams Kr mole 1 mole 22.4 Liters 83.8 / 22.4 = 3.74 g/L Kr 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 - % by mass of each element in a compound Percent = Part x 100 Whole Al: S: O: Percent Comp = Mass of 1 element x 100 Mass of compound Example: Find the mass percent composition of Al2(SO4)3 2 x 27 = 54 3 x 32 = 96 12 x 16 = 192 54 x 100= 15.8% % Al: 342 % S: 96 x 100 = 28.1% 342 %O: 192 x 100 = 56.1% 342 Total Comp. = 342 Percent Composition 18.8% Na # grams in 100 grams How many moles of each element? Divide by smallest #moles 29.0% Cl 52.2% O 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: either the same as the empirical formula (as for ionic compounds) or a simple whole-number multiple of the 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 40.1 g = 0.0224 mole Ca Cl: 1.60g 1 mole 35.5 g = 0.0451 mole Cl Step 2: DIVIDE the # of moles of each substance by the smallest number to get the simplest mole ratio. Ca: 0.0224 = 1 0.0224 Cl: 0.0451 = 2.01 ~ 2 0.0224 CaCl2 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 1. 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? 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 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? 58.8% C 1 mole C 12 g C = 4.9 / 1.9 = 2.5 x 2 = 5 9.8% H 1 mole H = 9.8 / 1.9 = 5 x 2 = 10 1gH 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 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: 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. 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. N2 (g) + 3H2 (g) 2NH3 (g) There is a 1:3:2 mole ratio 1. If you have 2 moles of N2, how many moles of NH3 will be produced? 2. If you want 5 moles of product, how many moles of hydrogen gas do you need? 3. How many moles of nitrogen are needed to react completely with 8 moles of hydrogen? N2 + 3H2 2NH3 If you have 2 moles of N2, how many moles of NH3 will be produced? 2 mol N2 2 mol NH3 1 mol N2 2 x 2 / 1 = 4 moles NH3 N2 + 3H2 2NH3 If you want 5 moles of product, how many moles of hydrogen gas do you need? 5 mol NH3 3 mol H2 2 mol NH3 5 x 3 / 2 = 7.5 moles 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 ****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!!!**** Note – If you don’t have moles already, your first step is to convert to moles! Mole Review: 22.4 L at STP 1 mole Molar Mass 6.02 x 1023 particles LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS Molar Mass MOLES (g/mol) 6.02 1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION NUMBER OF PARTICLES 1. 2. 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) N2 (g) + 3H2 (g) 2NH3 (g) 1. 2. 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? 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 17 g NH3 3 mole H2 2 g H2 2 mole NH3 1 mole H2 5.0 x 3 x 2 / 17 / 2 = 0.88 g H2 1) 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? 2) 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? 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) 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: 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. 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? **The amount of product that can be formed in a reaction is always determined by the limiting reactant!!** A s’mores MUST have: ◦ 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? 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) 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? _____________________ 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? _______________ 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 Theoretical Yield x 100% 1. 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? 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)