Stoichiometry Chapter 9 ©2011 University of Illinois Board of Trustees .

StoichiometryStoichiometry

Chapter 9Chapter 9

©2011 University of Illinois Board of Trustees • http://islcs.ncsa.illinois.edu/copyright

Do you remember?Do you remember? What is a Mole? A mole (mol) is an

amount of substance that contains the same number of particles as there are atoms in 12 g of carbon -12.

To four significant figures, there are 6.022 x 1023 atoms in 12 g of carbon-12.

Thus a mole of natural carbon is the amount of carbon that contains 6.022 x1023 carbon atoms.

The number 6.022 x 1023 is often called Avogadro’s number. ©2011 University of Illinois Board

of Trustees • http://islcs.ncsa.illinois.edu/copyrig

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StoichiometryStoichiometry

Greek stoicheion ( to measure the elements)

The proportional relationship between two or more substances during a chemical reaction

Branch of chemistry that deals with quantities of substances in chemical reactions.

©2011 University of Illinois Board of Trustees •

http://islcs.ncsa.illinois.edu/copyright

History of StoichiometryHistory of Stoichiometry

Joseph Black (1728-1799) and his student Daniel Rutherford (1749-1819) quantitatively studied the following reaction in both the forward and reverse direction to measure the redistribution of mass during the reaction:

CaCO3(s) → CaO(s) + CO2(g)

Their findings laid the groundwork for modern stoichiometry.

Daniel Rutherford



Balanced Equations Show Balanced Equations Show ProportionsProportions

A recipe for Kool-aid A recipe for Kool-aid 1 cup sugar1 cup sugar1 package of Kool-aid1 package of Kool-aid2 quarts of water2 quarts of waterHow do you make 4 How do you make 4

quarts?quarts?

A recipe for Water A recipe for Water 2H2H22 + O + O22 →→ 2H 2H22OO

The coefficients show the The coefficients show the relative amounts (ratio) in relative amounts (ratio) in equations. These can be equations. These can be expressed in moles, expressed in moles, molecules, or ions.molecules, or ions.



4 Questions4 Questions

There are 4 questions you need to ask yourself when solving problems, whether in real life or in class.

1.1. What do you know?What do you know?

2.2. Where do you want to Where do you want to go?go?

3.3. How do you get there?How do you get there?

4.4. Does it make sense?Does it make sense?©2011 University of Illinois Board


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Mole RatioMole Ratio is the Key to bridge the is the Key to bridge the gap when converting from one gap when converting from one

substance to another.substance to another. The The mole to mole ratiomole to mole ratio is represented by is represented by the coefficients in a balanced equation.the coefficients in a balanced equation.

NN22 + + 33HH22 →→ 22NHNH33

______mol knownmol known x x mol unknown mol unknown = ___mol unknown= ___mol unknown mol knownmol known Always start with what you know and Always start with what you know and

make sure the units cancel !!!make sure the units cancel !!!©2011 University of Illinois Board


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Sample Problem #1Sample Problem #1Mole to MoleMole to Mole

NN22 + + 33HH22 →→ 22NHNH33

What do we know? Where do we want to go? What do we know? Where do we want to go? 312 312

moles NHmoles NH33 (known) (known) _?__?_moles of Hmoles of H22 (unknown) (unknown)

How do we get there?How do we get there?

312 312 mol NHmol NH33 x x 33mol Hmol H22 = _____mol H= _____mol H22

22 mol NHmol NH33

Does it make sense?Does it make sense?

yesyes

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Getting into Moles and getting out Getting into Moles and getting out of Molesof Moles

Substances are usually measured Substances are usually measured by mass (or volume).by mass (or volume).

You may need to convert between You may need to convert between units for mass, volume, and mole.units for mass, volume, and mole.

Steps to convert unitsSteps to convert units1.1. Change units given into molesChange units given into moles2.2. Use Use mole ratiomole ratio to determine moles of to determine moles of

desired substance.desired substance.3.3. Change moles to whatever unit you Change moles to whatever unit you

need. need. ©2011 University of Illinois Board of Trustees •


Molar Mass for ElementsMolar Mass for Elements

Use Atomic Mass from the Periodic Table Use Atomic Mass from the Periodic Table for a conversion factor.for a conversion factor.

g elementsg elements

1mol element1mol element Molar Mass Calculation for CarbonMolar Mass Calculation for Carbon What do we know? Where do we want to go? What do we know? Where do we want to go?

0.55 g C0.55 g C ? ? mol Cmol C How do we get there?How do we get there?

0.55 g C x 0.55 g C x 1 mol C__ 1 mol C__ = 0.05 mol = 0.05 mol 12.01 g C12.01 g C

Does it make sense?Does it make sense?yesyes©2011 University of Illinois Board


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Sample Problem #2Sample Problem #2Mass to MassMass to Mass

NN22 + + 33HH22 →→ 22NHNH33 What do we know? Where do we want What do we know? Where do we want

to go? to go? 1221 g H1221 g H22 ? ? mass of NHmass of NH33

How do we get there?How do we get there?We use a We use a Mole to Mole Ratio ( from the balanced equation)Mole to Mole Ratio ( from the balanced equation)

and and Molar Mass Conversions (from the periodic table)Molar Mass Conversions (from the periodic table)

1221 g H1221 g H2 2 x x 1 mol H1 mol H22 x x 2 mol NH2 mol NH33 x x 17.04 g NH17.04 g NH33 = = __ g __ g NH3 2.02 g H2.02 g H22 3 mol H3 mol H22 1 mol 1 mol NHNH33

Does it make sense?Does it make sense?YesYes

Stoich Video Guest mass-massStoich Video Guest mass-mass

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Volume-Volume ProblemsVolume-Volume Problems

Conversions of volume to mass or Conversions of volume to mass or mass to volume usemass to volume use

1.1. Density as conversion factor for liquidsDensity as conversion factor for liquids2.2. Molar volume of a gas, for gases at STPMolar volume of a gas, for gases at STP3.3. Concentration of a solution, for an Concentration of a solution, for an

aqueous solutionaqueous solution Don’t forget the units you want to Don’t forget the units you want to

cancel should be on the bottom of cancel should be on the bottom of your conversion factor.your conversion factor.



Sample Problem #3Sample Problem #3Volume to VolumeVolume to Volume

POClPOCl33(l) + 3 H(l) + 3 H22O O →→ H H33POPO44(l) + 3 HCl(g)(l) + 3 HCl(g) What do we know?What do we know? Where do we want to Where do we want to

go?go? 56 mL POCl56 mL POCl33 ___mL ___mL HH33POPO44 1.67g/mL = density of1.67g/mL = density of POClPOCl3 3

153.32g/mol = molar mass of POCl153.32g/mol = molar mass of POCl3 3

1.83 g/mL = density H1.83 g/mL = density H33POPO4 4

98.00 g/mol = molar mass H98.00 g/mol = molar mass H33POPO44

How do we get there?How do we get there?56 mL POCl56 mL POCl33 x 1.67 g POCl x 1.67 g POCl3 3 x 1 mol POClx 1 mol POCl3 3 xx 1 mol H1 mol H33POPO4 4 xx 98.00 g98.00 g HH33POPO4 4 x 1 mL Hx 1 mL H33POPO4 4 = _____ mL H= _____ mL H33POPO4 4

1 mL POCl 1 mL POCl33 153.32 g POCl153.32 g POCl3 3 1 mol POCl1 mol POCl3 3 1 mol H1 mol H33POPO4 4 1.83 g H1.83 g H33POPO44

Does it make sense?Does it make sense?yesyes

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Particle ProblemsParticle Problems

Use Avogadro’s number as Use Avogadro’s number as conversion factor.conversion factor.

6.022 x 106.022 x 102323 particles particles

1mole1mole



Sample Problem #4Sample Problem #4

CC55HH12 12 (l) (l) →→ C C55HH8 8 (l) + 2H(l) + 2H22 (g) (g)

What do we know? Where do we want to go?What do we know? Where do we want to go?

1.89 x 101.89 x 102424 molecules C molecules C55HH1212 ? grams C ? grams C55HH88

6.022 x 106.022 x 102323 molecules/mol molecules/mol

68.13 g/mol molar mass C68.13 g/mol molar mass C55HH8 8

1 mol C1 mol C55HH12 12 = 1 mol C= 1 mol C55HH88

How do we get there?How do we get there?1.89 x 101.89 x 102424 molecules C molecules C55HH12 12 x x 1mol C 1mol C55HH1212________ x x 1 mol C1 mol C55HH8 8 x x 68.13g C 68.13g C55HH8 8 = _= _214214

g Cg C55HH88 6.022 x 106.022 x 102323 molecules molecules 1 mol C1 mol C55HH12 12 1 mol C 1 mol C55HH88

Does it Make sense?Does it Make sense? YesYes



Study GuideStudy GuideGeneral StepsGeneral Steps

Convert the given Convert the given unit to moles of the unit to moles of the first substance.first substance.

Convert moles of the Convert moles of the first substance to first substance to mole of the second mole of the second substance using the substance using the molar ratio derived molar ratio derived from the formula for from the formula for the compound.the compound.

Convert moles of the Convert moles of the second substance to second substance to the desired units of the desired units of the second the second substance.substance.



Limiting Reactants and Limiting Reactants and Percentage YieldPercentage Yield

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Chem Teachers Having Fun Chem Teachers Having Fun with Stoichwith Stoich



ReactantsReactants

Limiting Reactant Limiting Reactant ingredient that limits the amount of a ingredient that limits the amount of a

product that can form.product that can form. The reactant that is used up firstThe reactant that is used up first Often the most expensive ingredientOften the most expensive ingredient

Excess Reactant Excess Reactant Ingredient that there is more than enough Ingredient that there is more than enough

ofof Reactant that is left over after the reactionReactant that is left over after the reaction





YieldsYields

Theoretical YieldTheoretical Yield Maximum quantity of a product that a Maximum quantity of a product that a

reaction could theoretically makereaction could theoretically make Calculated based on limiting reactantCalculated based on limiting reactant The amount that should be made if it The amount that should be made if it

was perfect conditions, and a perfect was perfect conditions, and a perfect reaction. Stoichiometry is what reaction. Stoichiometry is what determines this amount. determines this amount.





Actual YieldActual Yield Determined experimentallyDetermined experimentally Mass of products actually formed usually Mass of products actually formed usually

less than expected (theoretical yield)less than expected (theoretical yield) Why?Why?

Many reactions don’t completely use limiting Many reactions don’t completely use limiting reactantreactant

Product often needs to be purified and some Product often needs to be purified and some is lost in this process.is lost in this process.

Other reaction (side reactions) are going on Other reaction (side reactions) are going on and using reactantsand using reactants



Percentage YieldPercentage Yield Ratio of actual yield to theoretical yield.Ratio of actual yield to theoretical yield.

___Actual yield___ ___Actual yield___ x 100 = percent yieldx 100 = percent yield Theoretical yieldTheoretical yield

Need more help check out this Need more help check out this website. website. bhs.smuhsd.org/.../briolgastoich.hbhs.smuhsd.org/.../briolgastoich.htmltml ©2011 University of Illinois Board


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Stoichiometry Chapter 9 ©2011 University of Illinois Board of Trustees .

Documents

mole mol

mol h2

mol unknown mol knownalways

x mol unknown

mol nh3does

mol nh3 x

mole of natural carbon

g of carbon