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The Mole Stoichiometry: Cookbook Chemistry

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The Mole A mole is a number Avogadro’s number = 6.02x10 23 Named after Amadeo Avogadro Loschmidt determined the number of particles in one cubic centimeter of a gas at ordinary temperature and pressure

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Counting atoms by counting moles By counting moles, atoms or molecules are counted Counting atoms by using moles eliminates waste in chemical reactions Coefficients in chemical equations represent mole quantities

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Counting atoms by counting moles 2Na + Cl 2 2NaCl “Two moles sodium and one mole chlorine gas react to give two moles sodium chloride” 4 moles sodium require 2 moles Cl 2 5.2 moles sodium require 2.6 moles Cl 2 3.1 moles Cl 2 require 6.2 moles sodium 2:1 is the sodium/chlorine mole ratio

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Counting atoms by counting moles Counting atoms allows prediction of product quantities 2Fe + 6HCl 2FeCl 3 + 3H 2 How many moles iron (III) chloride can be made using 4.3 moles HCl? Set up a proportion Coeff. 2mol FeCl 3 = x mol FeCl 3 prob. side 6mol HCl 4.3mol HCl side x=2(4.3)/6=1.43 mol FeCl 3

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Limiting reagents If mole quantities are not exact, one of the reactants will run out first – this reactant is the limiting reagent 2H 2 + O 2 2H 2 O If 3 moles H 2 are reacted with 1 mole O 2, what is the limiting reagent?

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Limiting reagents Divide each mole quantity by the coefficient to find equivalents. H 2 3/2=1.5eq O 2 1/1=1eq limiting reagent The reactant with the fewest equivalents (O 2 ) is the limiting reagent. The other (H 2 ) is “in excess”.

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Limiting reagents 22Fe + 6HCl 2FeCl 3 + 3H 2 00.0037 mol Fe is reacted with 0.017 mol HCl. What is the limiting reagent? FFe 0.0037/2 = 0.00185eq Fe HHCl 0.017/6 = 0.00283eq HCl

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Using limiting reagents The quantity of product obtained is limited by the amount of the limiting reagent 2H 2 + O 2 2H 2 O If 4.5 moles hydrogen gas and 1.9 moles oxygen are reacted, how many moles water will be formed?

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Using limiting reagents Solution: First determine the limiting reagent. H 2 : 4.5/2=2.25eq O 2 : 1.9/1=1.9eq limiting reagent Then set up a proportion between the limiting reagent and the desired product. O 2 1 = 1.9 H 2 O 2 x x=3.8 moles

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Molar Mass Molar mass is the mass of one mole of particles Atomic mass – found in the bottom of each square of the periodic table – units are grams/mole Atomic mass is the weighted average of the mass numbers of all the isotopes of an element. Molecular mass – the mass of one mole of molecules It is equal to the sum of the atomic masses of all the atoms in the molecule.

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Molar mass H 2 O – (H) 2x1 = 2 (O) 1x16 = 16total = 18g/mol NH 3 – (N) 1x14 = 14 (H) 3x1 = 3total = 17g/mol glucose (C 6 H 12 O 6 ) (C) – 6x12 = 72 (H) – 12x1 = 12 (O) – 6x16 = 96total = 180g/mol

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Molar mass Formula mass is the sum of all the atomic masses in a formula unit (for salts) NaCl – (Na) 23 (Cl) 35.5total = 58.5g/mol Mg(NO 3 ) 2 (Mg) 1x24.3 = 24.3 (N) 2x14 = 28 (O) 6x16 = 96 total = 148.3g/mol

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Using molar mass Mass to moles conversions mass/(molar mass) = moles g g/mol = g x mol/g = moles Example: How many moles are represented by 2.5 grams of water? Solution: 2.5g/(18g/mol) = 0.14mol

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Using molar mass Moles to mass conversions molesx(molar mass) = mass mol x g/mol = g Example: What is the mass of 0.094 moles sodium chloride? Solution: 0.094mol x 58.5g/mol = 5.5g

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Mass-mass stoichiometry 2HNO 3 + H 2 O 2 + 2Fe(NO 3 ) 2 2Fe(NO 3 ) 3 + 2H 2 O How many grams hydrogen peroxide (H 2 O 2 ) are needed to make 2.43 grams iron (III) nitrate (Fe(NO 3 ) 3 ) according to the reaction below? x g H 2 O 2 2.43 g iron (III) nitrate moles H 2 O 2 moles iron (III) nitrate mole ratio (2:1) by molar mass x by molar mass

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Per cent yield Mass obtained from calculations is “theoretical yield” – never obtained in practice Per cent yield = actual yield x 100% theoretical yield

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Per cent yield Jorma makes drugs for a hobby (aspirin, that is) and expects to obtain 2.13g aspirin from his synthesis reaction. In reality he only gets 1.89g. What is his % yield? (1.89/2.13)x100% = 88.7%

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Per cent composition by mass % composition by mass is a tool for compound identification To calculate: divide the molar mass contribution of each element by the total molar mass and multiply by 100%

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Per cent composition by mass Example: H 2 SO 4 (sulfuric acid) total molar mass H: 1x2=2 S: 32x1=32 O: 16x4=64sum=98g/mol %H=2(100%)/98=2.04% %S=32(100%)/98=32.65% %O=remainder=65.31%=64(100%)/98

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Determining formulas from % composition Formulas are a mole ratio of elements Empirical formula: simplest mole ratio of elements, like NaCl or Ca(NO 3 ) 2 Applies to any type of compound Molecular formula: mole ratio of elements in an actual molecule (all nonmetals), like H 2 O or NH 3 Often the molecular formula and the empirical formula are the same, but not always

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Determining formulas from % composition Hydrazine, a rocket fuel molecular formula – N 2 H 4 empirical formula – NH 2 Hydrogen peroxide molecular formula – H 2 O 2 empirical formula – HO Glucose, a sugar molecular formula – C 6 H 12 O 6 empirical formula – CH 2 O

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Determining formulas from % composition % composition is a mass ratio – so by converting mass to moles, the empirical formula can be determined. Example: Laboratory analysis finds a compound to consist of 28.05% Na, 29.27% C, 3.67% H, and 39.02% O. What is the empirical formula?

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Determining formulas from % composition Treat the % like grams Convert grams to moles Na: 28.05g/(23g/mol) = 1.22 mol C: 29.27g/(12g/mol) = 2.44 mol H: 3.67g/(1g/mol) = 3.67mol O: 39.02g/(16g/mol) = 2.44 mol

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Determining formulas from % composition Convert to simplest whole number ratio – divide all mol quantities by the smallest one. These results become the subscripts in the formula. Na: 1.22/1.22 = 1 C: 2.44/1.22 = 2 H: 3.67/1.22 = 3 O: 2.44/1.22 = 2 So the empirical formula is NaC 2 H 3 O 2 (sodium acetate).

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Ideal gas law Boyle’s Law: PV = C (P 1 V 1 = P 2 V 2 ) Factors that affect pressure/volume: Temperature (T) Amount (moles) of gas (Avogadro’s Principle) (n) Ideal Gas Law: PV nT Constant of proportionality = R (gas constant)

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Ideal Gas Law Ideal gas Law: PV = nRT V must be liters, T is Kelvins, n is moles Values for gas constant (depends on pressure units) P in atm: R = 0.08206Latm/molK P in kPa: R = 8.314LkPa/molK P in torr: R = 62.4Ltorr/molK

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Ideal Gas Law Example: Find the moles of oxygen in a balloon of 2.3L volume and 1.3atm pressure if the temperature is 45ºC. Solution: PV = nRT 1.3(2.3) = n(0.08206)(45+273) n = 1.3(2.3)/(0.08206)(45+273) n = 0.115 mol

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Ideal Gas Law Example 2: Find the molar volume of a gas at STP. Solution: STP = standard temperature and pressure (273K and 1 atm) PV = nRT 1V = 1(0.08206)(273) = 22.4L/mol

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