UNIT 6 STOICHIOMETRY 1
There are three ways to measure matter count (number of particles representative particles) mass (grams) volume (Liters) Mole unit for amt of matter relating these quantities 2
Representative Particles (rp) The type of particle depends on the type of substance elements atoms molecular cmpds (2 nonmetals) molecules ionic cmpds (metal + nonmetal) formula units (fu) Avogadro determined the # of any type of particle in one mole is constant Avogadro s # = 6.02 x 10 23 There are ALWAYS 6.02 x 10 23 particles in 1 mole of anything 6.02 x 10 23 rp = 1 mole 3
Conversion factor any fraction made up of equivalent measures conversion factors = 1 used to change from one unit to another Factor Label Method problem solving process that uses conversion factors Problems are set up so that units cancel one another out 4
To Convert from moles to rp or rp to moles: rp s are: atoms, molecules or fu 6.02 x 10 23 rp 1 mole 1 mole 6.02 x 10 23 rp Perform the following conversions: 25 moles Ag =? atoms Ag 25 moles 6.02 x 10 23 atoms 1 mole 9.2 x 10 23 fu NaCl =? moles = 1.51 x 10 25 atoms 9.2 x 10 23 fu 1 mole 6.02 x 10 23 fu = 1.53 moles 5
Molar Mass # of grams per mole of any substance determined using the atomic mass in the PT Add the masses of all the elements in a cmpd Used to convert: mass moles Au 197.0 g 1 mole NaCl Na = 23.0 Cl = 35.5 58.5 g 1 mole Cu(NO 3 ) 2 1 Cu x 63.5 = 2 N x 14.0 = 6 0 x 16.0 = 63.5 28.0 96.0 187.5 g 1 mole 6
mass (g) moles 3.2 moles Ca(OH) 2 =? grams 1 Ca 2 O 2 H x 40.1 x 16.0 x 1.0 = 40.1 = 32.0 = 2.0 74.1 g 1 mole 3.2 moles 74.1 g 1 mole = 237.1 g 7
250 g H 3 PO 4 =? moles 3 H 1 P 4 O x 1.0 x 31.0 x 16.0 = 3.0 = 31.0 = 64.0 98.0 g mass (g) moles 1 mole 250 g 1 mole 98.0 g = 2.6 mol 8
Molarity (M) Unit of concentration for solutions # moles 1 L Used to convert from moles to L or L to moles for solutions EX: How many moles in 1.5 L of a 0.1 M solution of CaBr 2? Rewrite: 0.1 moles 1 L 1.5 L 0.1 moles 1 L = 0.15 moles 9
EX: You need 15 moles of H 2 SO 4 in a 6 M solution. How many liters of solution will you have? Rewrite: 6 moles 1L 15 moles 1 L 6 moles = 2.5 L 10
Molar Volume the volume of 1 mole of a gas at STP STP = Standard Temperature and Pressure 0 C and 1 atm 22.4 L 1 mole EX: How many moles of H 2 gas in a 3.4 L balloon? 3.4 L 1 mole 22.4 L = 0.15 moles 11
Representative particles (rp) atoms, molecules, formula units 6.02 x 10 23 rp 1 mole mass (g) molar mass (PT) # g 1 mole Mole gas @ STP 22.4 L 1 mole Liquid Solutions Molarity (M) # moles 1 L Volume (L) 12
Empirical Formula tells the relative number of atoms of each element that a cmpd contains simplest formula smallest whole # ratio of the atoms in a cmpd does NOT necessarily indicate the actual # s of atoms present in each molecule 13
Percent to mass, mass to mole, divide by small, multiply til whole 14
Example: p. 90-91 73.9% Hg and 26.1% Cl 73.9 g Hg 1 mol Hg = 0.368 mol Hg /0.368 = 1 200.6 g Hg 26.1 g Cl 1 mol Cl = 0.735 mol Cl /0.368 = 2 35.5 g Cl HgCl 2 15
Sample 3.13: p. 92 40.92 g C 1 mol C = 3.407 mol C /3.406 = 1 x 3 = 3 12.01 g C 4.58 g H 1 mol H 1.008 g H = 4.54 mol H /3.406 = 1.333 x 3 = 4 54.50 g O 1 mol O 16.00 g O = 3.406 mol O /3.406 = 1 x 3 = 3 C 3 H 4 O 3 16
Molecular Formula actual formula for a cmpd whole number mulitple of the empirical formula To determine the whole number multiple: compare the empirical molar mass with the experimental molar mass Experimental molar mass Empirical molar mass multiply the empirical formula by the whole number to get the molecular formula 17
From sample exercise 3.13, the empirical formula for a cmpd is C 3 H 4 O 3. The experimentally determined molecular mass is 176 amu. Determine the molecular formula. Empirical molar mass: 3 C x 12 = 36 4 H x 1 = 4 3 O x 16 = 48 88 amu Multiple: 176 = 2 88 Molecular Formula: C 3 H 4 O 3 x 2 C 6 H 8 O 6 18
Percent Composition Percentage by mass contributed by each element in a substance Can be determined experimentally or using the chemical formula C 12 H 22 O 11 12 C x 12 = 144 /342 x 100% = 42.1% 22 H x 1 = 22 /342 x 100% = 6.4% 11 O x 16 = 176 /342 x 100% = 51.5% 342 amu To check: make sure that the % s all add to 100% 19
Reaction Stoichiometry Using a balanced eqn, if we know the mass of one reactant or product, we can determine the masses of every other substance in the rxn. The coefficients can be interpreted as the relative numbers of moles of each substance mole ratio 20
p. 97 grams of A grams of B use molar mass use molar mass moles of A use mole ratio moles of B 21
EX: How many moles of water are produced from 1.57 moles of O 2. 2H 2 + O 2 2H 2 O 1.57 mol x mol 1.57 mol O 2 2 mol H 2 O 1 mol O 2 = 3.14 mol H 2 O mole ratio 22
EX: Calculate the mass of CO 2 produced when 1.00 g of butane is burned. 2 C 4 H 10 + 13 O 2 8 CO 2 + 10 H 2 O 1.00 g x g 1.00 g C 4 H 10 1 mol C 4 H 10 8 mol CO 2 44 g CO 2 58 g C 4 H 10 2 mol C 4 H 10 1 mol CO 2 4 C = 48 10 H = 10 58 g/1mol C 4 H 10 1 C = 12 2 O = 32 44 g/1 mol CO 2 3.03 g CO 2 23
EX: How many grams of O 2 can be prepared from the decomposition of 0.0367 moles of KClO 3? 2 KClO 3 2 KCl + 3 O 2 0.037 mol x g 0.0367 mol KClO 3 3 mol O 2 2 mol KClO 3 32 g O 2 1 mol O 2 2 - O = 32 g/1 mol O 2 1.76 g O 2 24
Limiting Reactant (reagent) The reactant that is completely consumed in a chem rxn determines or limits the amt of product formed Excess Reactant (reagent) reactant that has some left over after the rxn 25
What are the limiting & excess reagents? 10 bottles & 12 lids 142 seniors & 100 airplane tickets 10 slices of bread & 7 slices of cheese 2 slices of bread are required for each sandwich 26
Balanced eqn: 2 H 2 + O 2 2H 2 O Moles: 2 mol 1 mol 2 mol Molar mass: 2 g/mol 32 g/mol 18 g/mol Mass: 4 g 32 g 36 g Rxn according 2 mol + 1 mol 2 mol to bal. eqn: 4 g + 32 g 36 g What will happen if 3 mol of H 2 are combined with 1 mol O 2? What will happen if 8 g of H 2 are combined with 100 g O 2? What will happen if 25 g of H 2 are combined with 125 g O 2? 27
To determine the limiting reactant, you must compare what you have to what you need. Sample Ex 3.19, p 101 2Na 3 PO 4 + 3Ba(NO 3 ) 2 Ba 3 (PO 4 ) 2 + 6NaNO 3 3.50 g 6.40 g x g x g 1 st 3.5 g Na 3 PO 4 1 mol Na 3PO 4 164 g Na 3 PO 4 3 mol Ba(NO 3 ) 2 2 mol Na 3 PO 4 261 g Ba(NO 3 ) 2 1 mol Ba(NO 3 ) 2 = 8.36 g Ba(NO 3 ) 2 needed to react with all the Na 3 PO 4 Since you have less than you need, Ba(NO 3 ) 2 is the limiting reactant and must be used for calculations. 28
2Na 3 PO 4 + 3Ba(NO 3 ) 2 Ba 3 (PO 4 ) 2 + 6NaNO 3 6.40 g x g 6.4 g Ba(NO 3 ) 2 1 mol Ba(NO 3) 2 261 g Ba(NO 3 ) 2 1 mol Ba 3 (PO 4 ) 2 3 mol Ba(NO 3 ) 2 602 g Ba 3 (PO 4 ) 2 1 mol Ba 3 (PO 4 ) 2 = 4.92 g Ba 3 (PO 4 ) 2 29
Another way: do comparison at the end 2Na 3 PO 4 + 3Ba(NO 3 ) 2 Ba 3 (PO 4 ) 2 + 6NaNO 3 3.50 g 6.40 g x g 6.4 g Ba(NO 3 ) 2 1 mol Ba(NO 3) 2 261 g Ba(NO 3 ) 2 1 mol Ba 3 (PO 4 ) 2 3 mol Ba(NO 3 ) 2 602 g Ba 3 (PO 4 ) 2 1 mol Ba 3 (PO 4 ) 2 = 4.92 g Ba 3 (PO 4 ) 2 3.5 g Na 3 PO 4 1 mol Na 3PO 4 164 g Na 3 PO 4 1 mol Ba 3 (PO 4 ) 2 2 mol Na 3 PO 4 602 g Ba 3 (PO 4 ) 2 1 mol Ba 3 (PO 4 ) 2 = 6.42 g Ba 3 PO 4 Limiting reactant: Ba(NO 3 ) 2 30
Theoretical yield the max. amt of product that can be produced from a given amt of reactant calculated amt of product from the limiting reactant Actual yield the measured amt of product obtained from a rxn almost always less than theoretical yield Percent yield ratio of actual yield to theoretical yield expressed as a % % yield = actual yield x 100% theoretical yield 31
Sample Ex 3.20 p 102 2C 6 H 12 + 5O 2 2H 2 C 6 H 8 O 4 + 2H 2 O 25 g x g 25 g C 6 H 12 1 mol C 6 H 12 84 g C 6 H 12 2 mol H 2 C 6 H 8 O 4 2 mol C 6 H 12 146 g H 2 C 6 H 8 O 4 1 mol H 2 C 6 H 8 O 4 Theoretical yield = 43.5 g H 2 C 6 H 8 O 4 Actual yield = 33.5 g % yield = 33.5 g x 100% 43.5 g = 77.0% 32