Presentation on theme: "The Mole Stoichiometry: Cookbook Chemistry. The Mole A mole is a number Avogadro’s number = 6.02x10 23 Named after Amadeo Avogadro Loschmidt determined."— Presentation transcript:
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
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
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
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
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?
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”.
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
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?
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
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.
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
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
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
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
Per cent yield Mass obtained from calculations is “theoretical yield” – never obtained in practice Per cent yield = actual yield x 100% theoretical yield
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%
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%
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
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
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
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?
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
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).
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)
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
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
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