“A” students work (without solutions manual) ~ 10 problems/night

“A” students work(without solutions manual)~ 10 problems/night.

Alanah FitchFlanner Hall [email protected]

Office Hours W – F 2-3 pm

Module #4Mass and Stoichiometry

mailto:[email protected]

FITCH Rules

G1: Suzuki is SuccessG2. Slow me down G3. Scientific Knowledge is ReferentialG4. Watch out for Red HerringsG5. Chemists are LazyC1. It’s all about chargeC2. Everybody wants to “be like Mike” (grp.18)C3. Size MattersC4. Still Waters Run DeepC5. Alpha Dogs eat first

Che

mis

try

Gen

eral

Particle Symbol

Neutron, n Proton, p Electron, e

10 e

11H01n

Massg1.6749285x10-24

1.6726231x10-24

1.093898x10-28

These numbers are inconvenientInvoke General Rule #5: Chemists Are LazyCreate some unit (not g) that is more useful (atomic mass unit)

1 1 6 6 0 5 3 8 7 3 1 0 2 4amu x g .

Invoke General Rule #3There must be some reference state

Amu

1.008671.007280.00055

11

1 26

1 2

amumass o f C a tom

11 9 9 2 6 1 0

1 2

2 6

am ux g

.

Why a relative reference state of 126C?

Represents a compromise between Physicists and Chemists

Chemists used a reference state of 16

8O as an abundant element everything (but gold) reacts with. Measured mass changes

Physicists used a reference state of 1

1H to measureRelative velocities of gas phase atoms

126C required

Least adjustmentsTo put bothPhysics and chemistryOn same scale

Chemists must haveStarted out chunkingRocks in fire

Rock a ir RockOs gfire

n s ,

Properties and MeasurementsProperty Unit Reference StateSize m size of earthVolume cm3 mWeight gram mass of 1 cm3 water at specified Temp

(and Pressure)Temperature oC, K boiling, freezing of water (specified Pressure)

1.66053873x10-24g amu mass of 1C12 atom/12

Expressing masses in terms of amu is more convenient

% Abundance atomic masses (amu)98.89 12.00001.11 13.00335

61 2C

61 3C

8 22 0 6 Pb

8 22 0 8 Pb

8 22 0 7 Pb

8 22 0 4 Pb

Two Examples we Examined already of STABLE isotopes

% Abundance atomic mass (amu)1.36 203.97323.6 205.974522.6 206.975952.1 207.9766

Relative

Relative

On the periodic table what is the reported atomic mass for lead?

Atomic mass= 207.2

a tom ic m ass am uaverage

iso tope

iso topeiso topes

%

1 0 0

On the periodic table what is the reported atomic mass for lead?

Atomic mass= 207.2

Where did this number come from?

a tom ic m ass amaverage

iso tope

iso topeiso topes

%

1 0 0

a tom ic m ass am uPb average, . 2 0 7 1 7 7 0 9

% Abundance atomic mass (amu) %x(amu)1.36 203.973 301.8823.6 205.9745 4860.998222.6 206.9759 4677.6553452.1 207.9766 10877.1761899.98 Weighted Sum 20717.70976

Average amu 207.1770976207.2

On the periodic table what is the reported amu for C?

amu=12.01

Where did this number come from?

a tom ic m ass am uaverage

iso tope

iso topeiso topes

%

1 0 0

a tom ic m ass am uC average, .1 2 0 1

% Abundance atomic masses (amu) %(amu)/10098.89 12.0000 11.86681.11 13.00335 0.144337185100 12.01113719

1 21

1 2

1

1 6 6 0 5 3 8 7 3 1 06 0 2 2 1 06

1 2 61 2

2 42 3

61 2g C

C atom

amu

amu

x gx C atom s

.

.

N mole x a tom sAvogadro 6 0 2 2 1 0 2 3.

1 21

1 2

1

1 6 6 0 5 3 8 7 3 1 0

1

6 0 2 2 1 016

1 2 61 2

2 4 2 3g CC atom

amu

amu

x g

m ole

x a tom sm ole

. .

16 0 2 2 1 0 2 3

mole

x a tom s.

General Rule #4: Chemists Are Lazy: make this number unique

( )x am u g X mole a tom sxelem en tX 1

Amedeo AvogadroItalian (Turin)1756-1856Conte di Quaregna e di Cerreto

How many particles are there?

Mole Latin mass=moles of stuff

Marie the Jewess, 300 Jabir ibn Hawan, 721-815

Galen, 170 Abbe Jean Picard1620-1682

Galileo Galili1564-1642

Daniel Fahrenheit1686-1737

Evangelista Torricelli1608-1647

Isaac Newton1643-1727

Robert Boyle, 1627-1691

Blaise Pascal1623-1662

Anders Celsius1701-1744

Charles Augustin Coulomb 1735-1806

Amedeo Avogadro1756-1856

TV Madscientist

John Dalton1766-1844

B. P. Emile Clapeyron1799-1864

Jacques Charles 1778-1850

Germain Henri Hess1802-1850

Fitch Rule G3: Science is Referential

William ThompsonLord Kelvin, 1824-1907

James Maxwell1831-1879

Johannes DiderikVan der Waals1837-1923

Justus von Liebig (1803-1873

1825-1898Johann Balmer

James Joule (1818-1889)

Johannes Rydberg1854-1919

Rudolph Clausius1822-1888

Thomas Graham1805-1869

Heinrich R. Hertz, 1857-1894

Max Planck1858-1947

J. J. Thomson1856-1940

Linus Pauling1901-1994

Werner Karl Heisenberg1901-1976

Wolfgang Pauli1900-1958

ElementHCOPb

Amu1.00812.0116.00207.2

Mass (g)1.00824.0248.00207.2

Moles (n)1231

# atoms6.022x1023

12.044x1023

18.066x1023

6.022x1023

Example

Properties and MeasurementsProperty Unit Reference StateSize m size of earthVolume cm3 mWeight gram mass of 1 cm3 water at specified Temp

(and Pressure)Temperature oC, K boiling, freezing of water (specified Pressure)

1.66053873x10-24g amu mass of 1C12 atom/12

6.022x1023 mole atomic mass of an element in grams

Molar Massin grams is numerically equal to the sum of the

masses (in amu) of the atoms in the formula

MM n amu a amu b amu z am ui i A B Zi

. . . . . .

Lead carbonate = PbCO3

principle component paintbefore WWIIMolar mass= unknown

Pb 1 207.2 207.2C 1 12.01 12.01O 3 16.00 48.00 267.21

267.2

Example. Lead carbonate is principle componentof white lead used in all white paints prior to WWII.Calculate the molar mass of lead carbonate

MM n amu a amu b am u z am ui i A B Zi

. . . . . .

amuElement Total amu#atoms

Example: Acetylsalicylic acid, C9H 8O4, is the active ingredient of aspirin.What is the mass in grams of 0.509 mol acetylsalicylic acid?

Acetylsalicylic acidC9H 8O4

Active ingredient of aspirin0.509 mol acetylsalicylic acidMass?

MMm

n


. . . . . .

MM 9 1 2 0 1 8 1 0 0 8 4 1 6 0 0. . .

MMg

mole1 8 0 1 5.

1 8 0 1 5 0 5 0 9 9 1 7. . .g

molem ole m g

MM n m

Mass %

% #elem en t a tom s o f tha t e lem en ta tom ic w eigh t o f e lem en t

m olar m ass o f com pound

1 0 0 %

Calcium carbonate, commonly called is used in many commercial products to relieve an upset stomach. It has the formula of CaCO3. Because lead ore bodies form by substitution of lead onto old coral reefs (calcite or calcium carbonate) some antacid materials have been tested for their lead composition. What are the mass percents of Ca, C, and O in calcium carbonate?

CaCO3

commercial productupset stomachmass percent Camass percent Cmass percent Oold coral reefslead in antacids

% #elem en t a tom s o f tha t e lem en ta tom ic w eigh t o f e lem en t

m olar m ass o f com pound

1 0 0 %

MM n amui ii

Ca 40.08 40.08C 12.01 12.01O 3(16.00) 48.00

100.09

Calcium carbonate, commonly called is used in many commercial products to relieve an upset stomach. It has the formula of CaCO3. What are the mass percents of Ca, C, and O in calcium carbonate? Because lead ore bodies form by substitution of lead onto old coral reefs (calcite or calcium carbonate) some antacid materials have been tested for their lead composition.

Ca 40.08 40.08C 12.01 12.013O 3(16.00) 48.00

100.09

%.

.Ca

14 0 0 8

1 0 0 0 91 0 0 %

1 Ca

%.

.C

11 2 0 1

1 0 0 0 91 0 0 %

%.

.O

31 6 0 0

1 0 0 0 91 0 0 %

% .Ca 4 0 0 4

% . .C 11 9 9 9 2 % 1 2 0 0 %

% . .O 4 7 9 5 6 8 3 8 8 5 4 7 9 6 %

Calcium carbonate, commonly called is used in many commercial products to relieve an upset stomach. It has the formula of CaCO3. What are the mass percents of Ca, C, and O in calcium carbonate? Because lead ore bodies form by substitution of lead onto old coral reefs (calcite or calcium carbonate) some antacid materials have been tested for their lead composition.

Check answer47.9612.0040.04100%

Very Important!

Are We Done?

Converting between g, moles, and number of particles

grams Moles Formula Units

Molar

Mass

Avogadro’s

Number

The Golden BridgeStoichiometry or #of molesMM

m

n

Mass=m Moles=n particles

nN partic lesA

Example. Lead carbonate is principle componentof white lead used in all white paints prior to WWII.Determine the number of moles of lead carbonate in a sample of 14.8 g lead carbonate


principle component paintbefore WWII14.8 g PbCO3

number of moles, n = unknown


. . . . . .

MMm

n

Pb 207.2 207.2C 12.01 12.01O 3(16.00) 48.00 267.21

267.2

nm

MM

Example. Lead carbonate is principle componentof white lead used in all white paints prior to WWII.Determine the number of moles of lead carbonate in a sample of 14.8 g lead carbonate


principle component paintbefore WWII14.8 g PbCO3

number of moles, n = unknown

Pb 207.2 207.2C 12.01 12.013O 3(16.00) 48.00 267.21

267.2

n mole 0 0 5 5 3 8 9 2 2 1.ngg

m ole

1 4 8

2 6 7 2

.

.n m ole 0 0 5 5 4.

Simplest Formula from Chemical Analysis

1. Mass scale is based on atomic number of C2. Mass scale is therefore proportional to number of

atoms or moles3 Convert mass to moles = whole units of atoms4. If moles are fractional this implies that atoms are

fractional NOT ALLOWED by chemistry (remember we chemists do not break up atoms - that is reserved forphysicists)

5. So multiply until we get whole number ratios

(Empirical)

Example A 25.00-g sample of an orange compoundcontains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g ofoxygen. Find the simplest formula

Total =25.00gK= 6.64gCr=8.84 gO=9.52 gunknown - simplest formulaorange compound

(Empirical)

CHECK THAT THE COMPOUND IS PURE!!

6.648.849.5225.00

This means that we have accounted for the total weight ofThe compound and we are confident in assigning weightRatios.

n MM m

nm

MM

)

Example : A 25.00-g sample of an orange compoundcontains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g ofoxygen. Find the simplest formula

6 6 4

3 9 1 01

0 1 6 9 8 0 1 7 0.

.. .

gKgK

mole

n m olK molK nm

MM

(Empirical) Sig fig

8 8 4

5 2 0 00 1 7 0 0 0 1 7 0

.

.. .

gC rgC r

m ol

n m olC r m olC r

9 5 2

1 6 0 00 5 9 5 0 0 0 5 9 5

.

.. .

gOgO

mol

n m olO molO

0.170molK0.170molCr0.595mol O

Simplest Formula from Chemical Analysis

1. Mass scale is based on atomic number of C2. Mass scale is therefore proportional to number of

atoms or moles3 Convert mass to moles = whole units of atoms4. If moles are fractional this implies that atoms are

fractional NOT ALLOWED by chemistry (remember we chemists do not break up atoms - that is reserved forphysicists)

5. So multiply until we get whole number ratios

(Empirical)

Example : A 25.00-g sample of an orange compoundcontains 6.64 g of potassium, 8.84 g of chromium, and 9.52 g ofoxygen. Find the simplest formula

0.170molK0.170molCr0.595mol O

0 1 7 0

0 1 7 0

1 0 0

1

.

.

.molK

molC r

m olK

molC r

0 5 9 5

0 1 7 0

3 5 0

1

.

.

.molO

molC r

m olO

molC r

Make this non-fractional, multiply x2

(Empirical)

0 1 7 0

0 1 7 0

1 0 0

1

.

.

.molC r

m olC r

m olC r

m olC r

2 0 0

2

. molK

molC r

7 0 0

2

. molO

molC r

2 0 0

2

. molC r

m olC r

7 O : 2K : 2Cr

Formula: Cations First

2K : 2Cr : 7 O

K2Cr2O7

Divide bySmallest #

Sugar of lead (lead acetate) was used from 0 A.D. to 1750s A.D. to “sweeten” ethyl alcohol (wine) and to prevent wine from going bad.

One (unproven) theory is that the Roman leaders were poisoned by excessive lead acetate consumption leading to the fall of Rome.

Context for the next problem: The Romans weren’t reallyDrunkards engaging in orgies

Lead acetate is easily made by boiling wine in a lead pot = sapa, a sweetner used in many Roman recipes.

A recipe of Apicius whichUses ½ cup sapa

Example : The compound that the Romans supposedly drank in excess is ethyl alcohol, which contains the elements carbon, hydrogen, and oxygen. When a sample of ethyl alcohol is burned in air, it is found that 5.00 g ethyl alcohol goes to 9.55 g CO2 plus 5.87g H2O

What is the simplest formula of ethyl alcohol?

Romans 5.00g ethyl alcoholOrgies 9.55 g CO2

Fall of empire 5.87 g H2Ointoxicating properties simplest formula?ethyl alcohol burned in airelements C, H, O

5.00g ethyl alcohol9.55 g CO2

5.87 g H2Osimplest formula?

elements C, H, OBurned in air

5 9 5 5 5 8 722 2g gCO gH Oto ta l C H O

O( ) . .

All the C and H in the sample is convertedIn air (contains O2) to CO2 and H2O

so g C in CO2 represents g C in originalethyl alcoholAnd g H in H2O represents g H original





elements C, H, O

n gCOmoleCO

amu gCO

moleC

moleCOC

2

2

2 2

1

1

n gCOmole CO

gCO

moleC

moleCOC

9 5 5

1 2 0 1 2 1 622

2 2

.( . ( ))

n gCOmole CO

gCO

moleC

moleCOC

9 5 5

4 4 0 122

2 2

.( .

n moleC 0 2 1 7.n

m

MMm

MM

1





elements C, H, O

n gH OmoleH O

amu gH O

moleH

moleH Oh

2

2

2 2

2

1

n gH Omole H O

gH O

moleH

moleH OH

5 8 7

1 6 0 0 2 1 0 0 8

22

2

2 2

.( . ( . ))

n gH Omole H O

gH O

moleH

moleH OH

5 8 7

1 8 0 1 6

22

2

2 2

.( .

n moleH 0 6 5 1 7.n

m

MMm

MM

1

n moleH 0 6 5 2.





elements C, H, ONow know:

mol C 0.217mol H 0.652

gO in original sample?

g g g gto ta l C H O

5 0 0 2 6 1 0 6 5 7. . .g g g gto ta l C H O

Need to know mol O.

How are we going to get it? g gam ugC

amugc COCO

2

2

g ggC

ggc CO

COC

9 5 5

1 2 0 1

4 4 0 12 6 1

2

2

..

..

3 sig fig

g gam ugH

amugH H OH O

2

2

g ggC

ggc CO

COC

5 8 7

2 1 0 0 8

1 8 0 1 60 6 5 7

2

2

.( . )

..

3 sig figgO 5 0 0 2 6 1 0 6 5 7. . .

gO 1 7 3 3.

gO 1 7 3.





elements C, H, O

n gOmoleO

amu gOO

n gOmole O

gOH

1 7 3

1

1 6 0 0.

( .

n m oleO 0 1 0 8 1 2 5.

nm

MMm

MM

1

n moleO 0 1 0 8.

gO 1 7 3.

5 0 0 2 6 1 0 6 5 7. . .g g g gto ta l C H O

gO 5 0 0 2 6 1 0 6 5 7. . .





elements C, H, O

molO=0.108molH=0.652molC=0.217

0 1 0 8

0 1 0 81

.

.

mol

m olO

O

C2H6O

0 6 5 2

0 1 0 86 0 3 7 6

.

..

mol

m olH

O

0 2 1 7

0 1 0 82 0 0 9 2

.

..

mol

m olC

O



Divide by smallest

Chemical Formulas

Write reactions

Relate to MassRatios andStoichiometry

PredictChange in amounts

Consider what happens when one part necessary for theReaction (recipe) is limiting

Mass Relations and Stoichiometry

Mass Relations In ReactionsWriting and Balancing Chemical Reactions

Lead Hand Side Right Hand Side

Reactants Productsnumber atoms A as reactants = number atoms A as productsnumber atoms Bas reactants = number atoms B as productsnumber atoms C as reactants = number atoms C as products...Number atoms Z as reactants = number atoms Z as products

Atoms are Conserved: no alchemy

Steps to Balance Reaction Equations

1. Write a “skeleton” equation with molecular formulas ofreactants on left, products on right

2 Indicate the physical state of the reactants and productsa. (g) for a gasb. (l) for a liquidc. (s) for a solidd. (aq) for an ion or molecule in water (aqueous) solution

3. Chose an element that appears in only one molecular formulaon each side of the equation

4. Balance the equation for mass of that element a. placing coefficients in front of the molecular formula

NOT by changing subscripts in the molecular formula5. Continue for the other elements6. The best answer is the one which is simplest whole-number

coefficients

Example: Lead ore, Lead sulfide (with the common name “galena”), was “calcined” by early metallurgists to form a leadoxide used to purify silver from other metals. Calcined means toburn in the presence of oxygen. At low temperatures the yellow lead oxide PbO, litharge, is formed. At higher temperaturesthe a red lead oxide, minium, is formed, Pb3O4. The sulfur is converted in the reaction to the gas SO2. Minium wasthe predominate source of red paints and pigments for illuminated bibles from which we derive the word miniature. Write a balancedchemical equation for the reaction of lead ore, galena with oxygen gas (O2) to form minium and SO2 gas. minium

PbS O 2 Pb O3 4 SO 2Burned in To make and

PbS O Pb O SO 2 3 4 2

galena

PbS

Pb3O4

PbS O Pb O SO 2 3 4 2

1S 1S

1Pb 3Pb

3 2 3 4 2PbS O Pb O SO

3Pb 3Pb

Subscript indicating numberPb within a single molecule

3S 1S

3 32 3 4 2PbS O Pb O SO

3S 3S

4O+(3x2)O=10 O2O

Count up total O

Number inFront indicatesNumber ofmolecules

3 5 32 3 4 2PbS O Pb O SO

4O+3x2O=10O

5x2O=10O

3 32 3 4 2PbS O Pb O SO

3 5 32 3 4 2PbS O Pb O SOs g s g( ) ( ) ( )

Be sure to indicate the physical state of the reactants and products

Sets up proportionalities BUT How doe we make useOf those proportionalities?

3 5 1 3 STOICHIOMETRY

From previous slide

4O+3x2O=10O

Chemical Formulas

Write reactions





Chemical Formulas

Write reactions





2 MAIN Strategies to solving these problems

Step wise

1. Easy conceptually = series of proportionalities

1. Longer time to compute2. Easy to lose track of units3. Leads to rounding errors

String wise

1. Not easy conceptuallywhen you are starting out

1. Shorter time to compute2. Easy to keep track of units

Will model this oneTo begin with

BUT will eventuallyOnly use this method!!!

3 5 32 3 4 2PbS O Pb O SOs g s g( ) ( ) ( )

Determine:a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts

1. Find moles of SO2 by Stoichiometry2. Find grams of SO2 from those moles

x

m ol

m olSO

molOO

g

g1 3 4

3

52

2

2.( )

( )

STOICHIOMETRY

xm olSO

molOmol m olSO

g

gO g

3

51 3 4 0 8 0 4

2

222

( )

( )( )

. .

m nMM

gSO molSOSamu Oamu

molSO2 22

0 8 0 42

.

( )

gSO molSOgSO

molSO2 22

2

0 8 0 46 4 0 7

.

.

gSO molSOgSO

molSO2 22

2

0 8 0 4

.

gSO molSO

gSO

molSO2 2

2

2

0 8 0 43 2 0 7 2 1 6 0 0

.. ( . )

gSO gSO2 25 1 5 .

gSO 2 5 1 5 1 2 2 8 .

x m olm olSO

molO

gSO

molSOg gO

g

g

1 3 4

3

5

6 4 0 75 1 5 1 2 2 8 5 1 5

2

2

2

2

2

..

. .( )

( )

3 5 32 3 4 2PbS O Pb O SOs g s g( ) ( ) ( )

Determine:a) The mass in grams of SO2 formed when 1.34 mol of O2 reacts

Alternative Strategy is to fold all calculations together

x m ol

m olSO

molO

g amu g am u

molSOO

g

g

s O

1 3 43

5

22

22

2 2

.( )

( )

x m olm olSO

molO

gSO

molSOO

g

g

1 3 4

3

52

2

2

2

2

.( )

( )

3 5 32 3 4 2PbS O Pb O SOs g s g( ) ( ) ( )

Determine:a) The mass in grams of SO2 formed when 1.34 mol of O2 reactsb) The mass in grams of O2 required to form 1.000 kg of Pb3O4

Stepwise calculations1. Find moles of Pb3O4 formed

molPb O form ed

kgPb Og

kg

gPb O

molPb O

3 4

3 4

3

3 4

3 4

11 0

nm

MMm nMM

molPb O form ed

kgPb Og

kg

gPb O

molPb O

3 4

3 4

3

3 4

3 4

11 0

3 2 0 7 2 4 1 6 0 0

( . ) ( . )

molPb O form ed

kgPb Og

kg

gPb O

molPb O

3 4

3 4

3

3 4

3 4

11 0

6 8 5 6

.

molPb O form ed3 4 1 4 5 8 5 7 6 4 2 9 1 4 5 9 . .

molPb O form ed3 4 1 4 5 9 .

3 5 32 3 4 2PbS O Pb O SOs g s g( ) ( ) ( )

Determine:a) The mass in grams of SO2 formed when 1.34 mol of O2 reactsb) The mass in grams of O2 required to form 1kg of Pb3O4

1. Find moles of Pb3O4 formed 2. Use Stoichiometry to find moles of O2

3. Find grams of O2 from those moles

m n MM

x m olOgO

molO

7 2 9 5 2

2

2

.

x

m olPb O

molO

molPb O1 4 5 9

5

3 4

2

3 4.

x m olO molO 1 4 5 9 5 7 2 9 52 2. .

x m olOgO

molOg

7 2 9 5

3 2 0 02 3 3 42

2

2

..

.

x kgg

kg

m ol

gPb O

Pb O

Pb O

11 0

3 4

3 4

3 4

3

3 5 32 3 4 2PbS O Pb O SOs g s g( ) ( ) ( )


Alternative string strategy

x g 2 3 3 3 7.

x kgg

kg

m ol

g

m ol

m ol

g

m olPb O

Pb O

Pb O

O

Pb O

O

O

11 0

6 8 5 6

5 3 2 0 03 4

3 4

3 4

2

3 4

2

2

3

.

.

x kg Pb O 13 4 x kg

g

kgPb O

1

1 03 4

3

x kgg

kg

m ol

g

m ol

m olPb O

Pb O

Pb O

O

Pb O

11 0 5

3 4

3 4

3 4

2

3 4

3

x kgg

kg

m ol

g

m ol

m ol

g

m olPb O

Pb O

Pb O

O

Pb O

O

O

11 0 5

3 4

3 4

3 4

2

3 4

2

2

3

3 5 32 3 4 2PbS O Pb O SOs g s g( ) ( ) ( )


c) The mass in grams of PbS required to react with 6.00 g of O2

1. Get moles of O2

2. Use Stoichiometry to get moles of PbS3. Get grams of PbS

xgO

gO

molO

molO

6 0 0

3 2 0 00 1 8 7 52

2

2

2

.

..

nm

MMm nMM

x m ol 0 11 2 5.x

m olO

molPbS

m olO0 1 8 7 5

3

52 2.

Stepwise method

x g 2 6 9.

x m olPbSgPbS

m olPbSg

0 11 2 52 3 9 2 7

2 6 9 1 7 8 7 5..

.

x m olPbSam u amu

molPbSPb S

0 11 2 52 0 7 2 3 2 0 7

.. .

x m olPbS M PbS 0 11 2 5.

3 5 32 3 4 2PbS O Pb O SOs g s g( ) ( ) ( )


c) The mass in grams of PbS required to react with 6.00 g of O2

x g 2 6 9.

x g 2 6 9 1 7.

x gO 6 0 0 2. x gOmolO

gO

6 0 0

3 2 0 022

2

..

x gOmolO

gO

molPbS

molO

6 0 0

3 2 0 0

3

522

2 2

..

x gOmolO

gO

molPbS

molO

gPbS

molPbS

6 0 03 2 0 0

3

5

2 3 9 2 72

2

2 2

..

.

Stringwise

Chemical Formulas

Write reactions




Rules for Limiting Reagent

1. Calculate the amount of product that would be formed if the first reactant were completely consumed

2. Repeat for the second reactant3. Choose the smaller of the two amounts. This is the

theoretical yield of the product. 4. The reactant that produces the smaller amount of the

product is the limiting reagent.5. The other reagent is in excess, only part of it is consumed.

Limiting ReagentsJustus von Liebig’s Law of the Limiting (1803-1873 Germany)

I have 1 dozen eggs, 2 packages of chocolate chips, and An entire carton of flour, an entire carton of sugar, a new bottleOf vanilla, and a new box of baking powder. The chocolate chip recipe calls for 2 eggs, 3 cups flour, 1 cup Sugar, 1 package chocolate chips, 1 tbsp vanilla, and 2 Tbsp baking powder. The recipe results in 36 Chocolate chip cookies.a. What is the limiting reagent?

1 2 3 1 1 2 3 6pkgch ips eggs cflour csugar tbspvan illa tb sppowder cookieshea t

Chips 1pkgEggs 2Flour 3cSugar 1cvanilla 1tbspPowder 2tbsp Yield 36 cookies

2pkg12 eggs>3cups>1cup>1tbsp>2tbspYield??

?

2

3 6

1pkg

cookies

pkg ?

2

3 6

17 2pkg

cookies

pkgcookies

?

1 2

3 6

2eggs

cookies

eggs ?

1 2

3 6

21 9 2eggs

cookies

eggscookies

Smallest yield = 72 cookiesLimiting reagent = chips

Expelled from grammar school for detonating an explosiveMade from chemicals obtained from his father’s business

While I wasn’t looking, my son snitched 4 cookies and ran off to Panama.

% .yie ld cookies

7 2 4

7 21 0 0 % 9 4 4 4 4 4 4 9 4 %

Actual Yield=Theoretical Yield

(most of the time)

%ex p

yie ldyie ld

yie lderim en ta l

theoretica l

1 0 0 %

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“A” students work (without solutions manual) ~ 10 problems/night