ימיכ תומיכ :הירטמואיכיוטס 04, 2010 · 2x 6.02x1023 atoms 6.02x1023 molecules 2x 6.02x1023 ... Calculate the number of grams of MgO produced from 0 ... what is

19.07.101 © Prof. Zvi C. Koren

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כימות כימי: סטויכיאומטריה


StoichiometryStoicheion + metron

(element) (measure)

Weight relations in chemical rxns. based on conservation of matter

Examples:

2 “molecules”1 molecule2 atoms

2x 6.02x1023 “molec.”6.02x1023 molecules2x 6.02x1023 atoms

2 moles of “molecules”1 mole of molecules2 moles of atoms

80.6 g32.0 g48.6 g

2Mg(s) + O2(g) 2MgO(s)

For any rxn.,

The absolute value of each coefficient is meaningless by itself!

BUT, the RATIOS are HOLY!!!

IFThen: אז-אםrelationship

כימות כימי


Stoichiometric Calculations: The Approach

gram mole mole mole

moleA gram gram gramA

MW (or AW)

2. Think in Moles !!!

3. Setup a Flow-Chart whereby g mol mol

Helpful Tips for Solving Problems:

4. Always include Units and Substance Name

Simple formula: MW

mn

moleB Stoichiometry

(rxn)

1. Write the Balanced rxn!!


2Mg(s) + O2(g) 2MgO(s)

Stoichiometric Calculations: Examples

Calculate the number of grams of MgO produced from 0.145 g Mg.

grams

moles moles

grams ?

AW, MW MW

Stoichiometric Ratio:Rxn

0.145 g Mg

Mg 3050.24

Mg 1

g

mol

Mg 2

MgO 2

mol

mol

MgO 1

MgO 0.30444

mol

g= 0.240 g MgO

MW factor MW factorRxn factoror

Stoichiometricfactor

So, Remember, All Roads Go Through Moles !!!


The Limiting Reagent (or Limiting Reactant)

CCl4(l) + SbF3(s) CCl2F2(l) + SbCl3(s)23 3 2

• Limits quantities reacting and produced• Completely reacts (if rxn goes to completion)

150. g CCl4

4

4

CCl 53.8231

CCl 1

g

mol= 0.975 mol CCl4.

150. g CCl4 and 100. g SbF3 are in a flask. How many grams of CCl2F2 are produced?

1. Calculate the number of moles of each reactant:

100. g SbF3

3

3

SbF 78.74521

SbF 1

g

mol= 0.559 mol SbF3

2. Assume all of one reactant completely reacts and determine how much of the other is needed.

If all of CCl4 reacts, how many moles of SbF3 are needed?

IF 0.975 mol CCl4

4

3

CCl 3

SbF 2

mol

mol= 0.650 mol SbF3 needed. BUT, we DON’T have it!!!

IF 0.559 mol SbF3

3

4

SbF 2

CCl 3

mol

mol= 0.839 mol CCl4 needed. AND, we DO have it!!!

SbF3 is the limiting reagent!(CCl4 is the “reagent in excess”)

If all of SbF3 reacts, how many moles of CCl4 are needed?

3. Identify the Limiting Reagent:

4. Calculate all quantities based on the limiting reagent:

0.559 mol SbF3

3

22

SbF 2

FCCl 3

mol

mol

22

22

FCCl 1

FCCl 9138.120

mol

g= 101 g CCl2F2


Analyzing Mixtures of Similar Compounds

For example, consider NaHCO3(s) and Na2CO3(s). Both react with HCl(aq):

NaHCO3(s) + HCl(aq) NaCl(aq) + H2CO3(aq) H2O(l) + CO2(g)

Na2CO3(s) + HCl(aq) NaCl(aq) + H2CO3(aq) H2O(l) + CO2(g)

Problem:Calculate the percent of Na2CO3 in the mixture when a 10.0-g mixture of sodium bicarbonate and sodium carbonate react with an excess of hydrochloric acid to produce 8.3818 g of NaCl, collected after evaporation.Solution:Let the wonders of algebra solve all of your problems:

Do NOT combine the two rxns.!!!

Let x = grams of Na2CO3 in mixture

10.0 – x = grams of NaHCO3 in mixture

nNaCl,total = (nNaClbicarbonate) + (nNaClcarbonate)

= (nbicarbonate) + 2 • (ncarbonate)

Also simplify by using n = m/MW

3NaHCO of

0.10

MW

x

32CONa of 2MW

x

NaCl of

3818.8

MW

x = 3.50 g % Na2CO3 in mixture = (3.50 g/ 10.0 g) 100 = 35.0 %

= +

2 2


Percent Yield

% Yield = 100 x

yieldltheoretica

yieldactual

Many rxns do NOT go to completion. There is a chemical energy barrier involved.

C7H6O3(s) + C4H6O3(l) C9H8O4(s) + H2O(l)For example:

salicylicacid

aceticanhydride

acetylsalicylic

acid

Aspirin

If from 14.43 g of the acid, 6.26 g of aspirin is produced, what is the % yield for the rxn?

2 2

Theoretical yield or maximum yield (assuming rxn goes to completion):

cid 38.12261

cid 1

ag

amol

acid 2

aspirin 2

mol

mol

aspirin 1

aspirin 80.15981

mol

g= 18.82 g aspirin

% yield = 100 x 82.18

26.6

g

g= 33.3 %

14.43 g acid


Solution Concentrations

Molarity, Formality, molality, Normality, % w/w, % w/v

Molarity = M = n

/V# of moles of solute per liter of solution

# of mmoles of solute per mL of solution

moles/L

mmoles/mL

3.0 M

KMnO4

“3.0 molarpotassiumpermanganatesolution”

3.0 moles of KMnO4 per L of solution

3.0 mmoles of KMnO4 per mL of solution

Dilutions

Prepare 500.0 mL of a 0.100-M KMnO4 solution from a 3.0-M stock solution?

Use n = M•V

nsolute in new solution = Mnew • Vnew = (0.100 mol KMnO4/L)(0.5000 L) = 0.0500

= nsolute from stock solution = Mstock • Vstock = (3.0 mol KMnO4/L) • Vold

= n =Mstock • VstockMnew • Vnew

(3.0 M) • Vstock = (0.100 M) • (0.5000 L) Vstock = 0.0167 L = 16.7 mL

Preparation of a Diluted Solution:

• Remove 16.7 mL of the 3.0-M stock solution

• Add enough water (“483.3 mL”) to produce 500. mL of solution


Solution Stoichiometry

Titrations: Acid-Base & Redox

Buret

NaOH(aq) + H2C2O4(aq)

Acid-Base Titrations

Na2C2O4(aq) + H2O(l)2 2

Erlenmeyer

flask

Add indicator (or use pH meter) to indicate when the end point of

the rxn has been reached: color changes (base acid).

End-Point or Equivalence Point of an Acid-Base or Redox rxn: Point at which all of of the Acid reacts with all of the Base orall of the oxidant reacts with all of the reductant.

Problem 1: Reaching an end-point

How many mL of 0.300 M sodium hydroxide are needed to

titrate 25.0 mL of 0.400 M oxalic acid?

Solution:

n = M ·V

MA,VA moles acid moles base V baserxn V=n/M

nACID = MA ·VA =

Stoichiometric Flow-Chart:

VBASE = nBASE / MBASE = 0.0200 mol / 0.300 M = 0.0667 L = 66.7 mL

(0.400 M)(0.0250 L) = 0.0100 mol

nBASE = 2 • nACID = 2 • (0.0100 mol) = 0.0200 mol

n=M·V

Write the Rxn.:


Acid-Base Problems (continued)

Problem 2: Standardization of a solution

If 0.250 g of solid sodium carbonate requires 25.76 mL of hydrochloric acid for

titration to the equivalence point, what is the molar concentration of HCl?

Na2CO3 + HCl “H2CO3” + NaCl2 2

Solution:

Stoichiometric Flow-Chart:

g salt/base moles salt/base moles acid M acidrxn M=n/V

nNa2CO3 = m / MW =

32

32

CONa / 98874.105

CONa .2500

molg

g= 0.0023587 mol base

nHCl = 0.0023587 mol base

base mol 1

acid mol 2= 0.0047174 mol acid

MHCl = n / V = (0.0047174 mol) / (0.02576 L) = 0.183 M HCl

Write the Rxn.:

MW

basic salt acid


Redox Problems

Redox Titrations

(same stoichiometric principles as acid-base problems)

Problem:

In an acidic solution, 1.026 g of an iron(II)-containing ore requires 24.34 mL of

0.0200 M KMnO4 to reach the equivalence point, what is the weight percentage of

iron in the ore? [The two redox products are Fe3+ and Mn2+.]

Solution:

MnO4– + 8H+ + 5Fe2+

Mn2+ + 5Fe3+ + 4H2O

nKMnO4 = M • V = (0.0200 M)(0.02434 L) = 4.868 x 10–4 mole MnO4–

-

4

2

MnO 1

Fe 5

mol

mol

2

2

Fe 1

Fe 5.8475

mol

g= 0.136 g Fe2+

purple colorless colorless colorless

4.868 x 10-4 mole MnO4- •mFe2+ =

% Fe in ore = 100 x ore 026.1

Fe 136.0 2

g

g= 13.3 % Fe in ore

MKMnO4,VKMnO4 moles MnO4– moles Fe2+

grams Fe2+

Write the Rxn.: MnO4– + Fe2+ ----> Mn2+ + Fe3+

Stoichiometric Flow-Chart:rxnn=MV MW


Solution Concentrations (continued)

Molarity, Formality, molality, Normality , % w/w, % w/v

Normality = N = eq/V

# of equivalents of solute per liter of solution

# of meqs of solute per mL of solution

eqs/L

meqs/mL

1 equivalent of a substance is that quantity that reacts with (or produces)

1 mole of H+ (in acid-base rxns.) or 1 mole of e– (in redox rxns.), etc.

In general:

1 eq 1 mol transferred species (צּורֹון)

In addition:

Equivalent Weight (EW) = Weight (Mass) of 1 eq: “g/eq”

Recall: Molecular Weight (MW) = Weight (Mass) of 1 mol: “g/mol”

EW vs. MW# g/eq = (# g/mol)(1 mol/# eqs) EW MW

N vs. M # eq/L = (# mol/L)(# eqs/mol) N M

Koren’sAlphabet

Rule

In general, MW & M are absolute; EW & N are relative (depend on rxn.)


Examples:

H2SO4 as an acid (full rxn): 1 mol H2SO4 2 mol H+

1 mol H2SO4 2 eqs H2SO4

MW (H2SO4) = 98 g/mol.

EW (H2SO4) = (98 g/mol)(1 mol/2 eqs) = 49 g/eq

For a 3.0 M H2SO4 solution Normality = (3.0 mol/L)(2 eqs/mol) = 6.0 N---------------------------------------------------------------------------------------------------------------------------------------------------------

Na2Cr2O7 reduced to CrCl3: Na2Cr2O7 + 6e– ---> 2CrCl3:

1 mol Na2Cr2O7 6 mol e–

1 mol Na2Cr2O7 6 eqs Na2Cr2O7

MW (Na2Cr2O7) = 262 g/mol.

EW (Na2Cr2O7) = (262 g/mol)(1 mol/6 eqs) = 43.7 g/eq

For example, for a 3.0 M Na2Cr2O7 solution Normality = ____ N---------------------------------------------------------------------------------------------------------------------------------------------------------

Ca(OH)2: Ca(OH)2(aq) Ca2+ + 2OH– 1 mol Ca(OH)2 2 eq Ca(OH)2

H2O oxidized to H2O2: 2H2O H2O2 + 2e–

1 mol H2O 1 eq H2O

Normality (continued)


Normality & Titrations

In general, for chemical rxns between two reactants:

n1 n2 M1V1 M2V2

For example: H2SO4(aq) + 2NaOH(aq) Na2SO4(aq) + 2H2O(l)

But, the following is ALWAYS true (for acid-base or redox rxns, etc.):

# eqs1 = # eqs2 N1·V1 = N2·V2

Acid-Base Example:

Question: In the above acid-base rxn, what is the normality and molarity of the acid if

20.0 mL of the acid are titrated with 40.0 mL of 0.200 M base to reach the end-point?

Answer: For NaOH: N = M.

NA · VA = NB · VB

NA(20.0 mL) = (0.200 N)(40.0 mL) NA = 0.400 N MA = 0.200 M

Redox Example:

For the rxn: MnO4– + 8H+ + 5Fe2+

Mn2+ + 5Fe3+ + 4H2O

2244

2244 FeFeMnOMnOFeFeMnOMnO

M . N , M . N VN VN vsvs

Koren’s “Jealousy” Equation:


Terminology for dissolution: a solute is dissolved by a solvent to form a solution.

UnitsFormulaDefinitionName

MM = molssolute/Lsol’n# of moles of solute per L of solutionMolarity

NN = eqssolute/Lsol’n# of equivalents of solute p. L of sol’nNormality

g/mLdsol’n = msol’n/Vsol’nmass of the sol’n per volume of sol’nDensity of Sol’n

(d or , rho)

mm = molssolute/kgsolvent

|m| |M|

# of moles of solute per kg of solvent

In dilute aqueous solutions:Molality

(none)

%

Xi = ni/nsol’n, ΣXi = 1

Xi 100

moles of a component p. total moles

mole fraction of a comp. as a percent

Mole fraction

Mole percent

%w/w % = mi/msol’n100

w/v % = gi/mLsol’n100

weight of solute/weight of sol’n, as %

weight of solute/volume of sol’n, as % Weight percent

ppmppm = gi/gsol’n= mg/kg

ppm |mg/L|

# of solute parts p. million sol’n parts

In dilute aqueous solutions:Parts per million

ppbppb = ngi/gsol’n= g/kg

ppb |g/L|

# of solute parts p. billion sol’n parts

In dilute aqueous solutions:Parts per billion

%v/v %=Vi,pure/Vsol’n100vol. of pure solute/vol. of sol’n, as %Volume percent

ProofProof = 2(v/v %)Double the Volume % (for whiskey)Proof

More Solution Concentrations

Note: Must always write the units and the substance, e.g., 2.0 g solute.

Questions: What is ppt? ___________________ pph? ________________________


Selected Concentration Examples:

(1) 1.2 kg ethylene glycol (HOCH2CH2OH), an antifreeze, is added to 4.0 kg

water. Calculate (for ethylene glycol): mole fraction, molality, weight/weight%.

[Answers: X = 0.080, m = 4.8 m, w/w % = 23 %]

(2) 560 g NaHSO4 are dissolved in 4.5x105 L water at 25 oC. Calculate the Na+

concentration in parts per million. [Answer: 0.24 ppm]

(3) 10.0 g of sucrose (C12H22O11) are dissolved in 250. g of water. Calculate (for

sugar): X, m, w/w %. [Answers: X = 0.00210, m = 0.117 m, w/w % = 3.85 %.]

(4) Sea water has a sodium ion concentration of 1.08 x 104 ppm. If the Na is

present in the form of dissolved sodium chloride, how many grams of NaCl

are in each liter of sea water? (Density of sea water is 1.05 g/mL.)

[Answer: 28.7 g NaCl/L]

(5) A 0.100-M aqueous solution of ethylene glycol has a density of 1.09 g/mL.

What is the molality of the solution? [Answer: 0.0923 m]

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ימיכ תומיכ :הירטמואיכיוטס 04, 2010 · 2x 6.02x1023 atoms 6.02x1023 molecules 2x 6.02x1023 ... Calculate the number of grams of MgO produced from 0 ... what is